Linear Simultaneous Equations

Let take my sister’s present age to be = X

My father’s present age as = Y

Given that

The sum of my elder sister’s present age and my father’s present age is 55 years.

So we can write it as

Sister’s age + father’s age =55

X+Y = 55 ………… (1)

Also given that after 16 years, my father’s age will be two times of my sister’s age.

So father’s age after 16 years will be (Y+16)

And sister’s age after 16 years will be (X+16)

So our new equation become

Father’s age = 2× (sister’s age)

Y+16 = 2×(X+16)

Y = 2X +16 ………… (2)

Equation (1)

X +Y=55

Equation (2)

Y = 2X +16

After plotting the lines, we can determine the intersection point of lines.

You can clearly see in the figure given below.

Red line for equation (1)

Blue line for equation (2)

We can satisfy the point (13, 42)

Equation (1)

X+Y=55

13 +42 = 55

AND EQUATION (2)

Y=2X+16

42 =2×13+16

HENCE, sister’s age is =13 years

Father’s age is =24 years

Let’s take the value of pen = Rs X

Value of pencil = Rs Y

Given that

Mita has bought 3 pens and 4 pencils for Rs.42

So we can write the equation as

3(PEN) +4(PENCIL) =42

3X + 4Y =42 ……… (1)

Also given that

I bought 9 pens and 1 dozen pencils at the same rate to give gifts to my friends at the Rs. 126.

Equation become

9X+12Y =126 ……… (2)

If we divide both sides by 3 we get

3X +4Y = 42

Both equation (1) and (2) are the same, so there exist an infinite number of solutions.

We can plot this equation, see the figure given below -

Every point laying on this line on 1^st quadrant can be our answer …

So there is an infinite number of solution to this question.

Let’s take the price of one art papers to be = Rs X

And the price of one sketch pen be = Rs Y

Given that

I have bought 2 art papers and 5 sketch pens at Rs. 16

So the equations will become

2X+5Y =16 ………… (1)

Also given that

Dola has bought 4 art papers and 10 sketch pens of the same rate.

So the equation become

4(art papers) +10(pens) =16

4 X + 10Y = 16………… (2)

We can plot this line on graphs

Equation (1)

Equation (2)

We can see the plots given below –

As we can see that both lines are parrell to each other

So there is no solution to this equations

We cannot find any value of the price of pens and art papers.

2x + 3y – 7 = 0 y = ... equation (i)

Equation (i) will be plotted as line AB.

3x + 2y – 8 = 0 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here, we can see in the graph that lines AB and CD are intersecting at point A(2,1).Hence it is solvable.

Solution of equations is x=2 and y=1.

4x - y = 0 y = 4x ... equation (i)

Equation (i) will be plotted as line AB.

-8x + 2y = -22 y = .. equation (ii)

Equation (ii) will be plotted as line CD.

Here, we can see from graph that lines AB and CD are not intersecting each other i.e, they are parallel. Hence it is not solvable.

7x + 3y = 42 y = ... equation (i)

Equation (i) will be plotted as line AB.

21x + 9y = 42 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here, we can see from graph that lines AB and CD are not intersecting each other i.e, they are parallel. Hence it is not solvable.

5x + y = 13 y = 13 – 5x ... equation (i)

Equation (i) will be plotted as line AB.

5x + 5y = 12 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here, we can see in the graph that lines AB and CD are intersecting at point P(2.65,-0.25).Hence it is solvable.

Solution of equations is x=2.65 and y=-0.25.

x + 5y = 7 …(1)

x + 5y = 20 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

x + 5y = 7 x + 5y = 20

∴ x + 5y + (-7) = 0 ∴ x + 5y + (–20) = 0

Or 1 × x + 5 × y + (-7) = 0 or, 1 × x + 5 × y + (-20) = 0

Here a₁ = 1, b₁ = 5, c₁ = -7 and a₂ = 1, b₂ = 5, c₂ = -20

Comparing the ratio of , we get

, and

Here . Therefore, it is not solvable. Lines will be parallel.

Now, plot the lines on graph,

x + 5y = 7 y = ... equation (i)

Equation (i) will be plotted as line AB.

x + 5y = 20 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

We can also see clearly in the graph that two lines are not intersecting each other at any point.

2x + y = 8 …(1)

2y – 3x = –5 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

2x + y = 8 2y – 3x = –5

∴ 2x + y + (-8) = 0 ∴ 2y + (-3x) + 5 = 0

Or 2 × x + 1 × y + (-8) = 0 or, -3 × x + 2 × y + 5 = 0

Here a₁ = 2, b₁ = 1, c₁ = -8 and a₂ = -3, b₂ = 2, c₂ = 5

Comparing the ratio of , we get

, and

Here . Therefore, it is solvable and has one common solution. Lines will intersect at a point.

Now, plot the lines on graph,

2x + y = 8 y = 8 – 2x ... equation (i)

Equation (i) will be plotted as line AB.

2y – 3x = –5 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph also, we can see that two lines are intersecting each other at point P(3,2). So it has a common solution which is x = 3 and y = 2.

5x + 8y = 14 …(1)

15x + 24y = 42 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

5x + 8y = 14 15x + 24y = 42

∴ 5x + 8y – 14 = 0 ∴ 15x + 24y – 42 = 0

Or 5 × x + 8 × y + (-14) = 0 or, 15 × x + 24 × y + (-42) = 0

Here a₁ = 5, b₁ = 8, c₁ = -14 and a₂ = 15, b₂ = 24, c₂ = -42

Comparing the ratio of , we get

, and

Here . Therefore, it is solvable and has infinite common solutions. Lines will overlap.

Now, plot the lines on graph,

5x + 8y = 14 y = ... equation (i)

Equation (i) will be plotted as line AB.

15x + 24y = 42 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph also, we can see that both lines are overlapped on each other. Therefore it has infinite common solutions.

3x + 2y = 6 …(1)

12x + 8y = 24 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

3x + 2y = 6 12x + 8y = 24

∴ 3x + 2y + (-6) = 0 ∴ 12x + 8y + (-24) = 0

Or 3 × x + 2 × y + (-6) = 0 or, 12 × x + 8 × y + (-24) = 0

Here a₁ = 3, b₁ = 2, c₁ = -6 and a₂ = 12, b₂ = 8, c₂ = -24

Comparing the ratio of , we get

, and

Here . Therefore, it is solvable and has infinite common solutions. Lines will overlap.

Now, plot the lines on graph,

3x + 2y = 6 y = ... equation (i)

Equation (i) will be plotted as line AB.

12x + 8y = 24 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph also, we can see that both lines are overlapping on each other. Therefore it has infinite common solutions.

5x + 3y = 11 …(1)

2x – 7y = –12 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

5x + 3y = 11 2x – 7y = –12

∴ 5x + 3y + (-11) = 0 ∴ 2x – 7y + 12 = 0

Or 5 × x + 3 × y + (-11) = 0 or, 2 × x + (-7) × y + 12 = 0

Here a₁ = 5, b₁ = 3, c₁ = -11 and a₂ = 2, b₂ = -7, c₂ = 12

Comparing the ratio of , we get

, and

Here . Therefore, it has one common solution. Graph of equations will intersect at a point.

6x – 8y = 2 …(1)

3x – 4y = 1 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

6x – 8y = 2 3x – 4y = 1

∴ 6x + (-8y) + (-2) = 0 ∴ 3x + (-4y) + (-1) = 0

Or 6 × x + (-8) × y + (-2) = 0 or, 3 × x + (-4) × y + (-1) = 0

Here a₁ = 6, b₁ = -8, c₁ = -2 and a₂ = 3, b₂ = -4, c₂ = -1

Comparing the ratio of , we get

, and

Here . Therefore, it has infinite common solutions. Graph of equations will overlap.

8x – 7y = 0 …(1)

x + 5y = 20 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

8x – 7y = 0 8x -7y = 56

∴ 8x + (-7y) + 0 = 0 ∴ 8x + (-7y) + (-56) = 0

Or 8 × x + (-7) × y + 0 = 0 or, 8 × x + (-7) × y + (-56) = 0

Here a₁ = 8, b₁ = -7, c₁ = 0 and a₂ = 8, b₂ = -7, c₂ = -56

Comparing the ratio of , we get

, and

Here . Therefore, graph of equations will be parallel.

4x – 3y = 6 …(1)

4y – 5x = –7 …(2)

Let us express the equations (1) and (2) in the form of

ax + by +c = 0 where a and b can’t be 0 at the same time.

[In the first equation, we use a₁, b₁, c₁ and in second equation, we use a₂, b₂, c₂ ]

4x – 3y = 6 4y – 5x = –7

∴ 4x + (-3y) + (-6) = 0 ∴ –5x + 4y + 7= 0

Or 4 × x + (-3) × y + (-6) = 0 or, -5 × x + 4 × y + 7 = 0

Here a₁ = 4, b₁ = -3, c₁ = -6 and a₂ = -5, b₂ = 4, c₂ = 7

Comparing the ratio of , we get

, and

Here . Therefore, it has one common solution. Graph of equations will intersect at a point.

4x + 3y = 20 y = ... equation (i)

Equation (i) will be plotted as line AB.

8x + 6y = 40 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here we can see in the graph that, lines AB and CD overlaps on each other. Therefore, it has infinite number of solutions.

Among them 3 solutions are: x = 2 & y = 4

x=5 & y = 0

x= -1 & y=8

4x + 3y = 20 y = ... equation (i)

Equation (i) will be plotted as line AB.

12x + 9y = 20 y = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph, we can see that these two lines are parallel to each other. Hence it is not solvable.

4x + 3y = 20 y = ... equation (i)

Equation (i) will be plotted as line AB.

6x – y = 8 y = 6x - 8 ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph, we can see that these two lines are intersecting each other at point P(2,4). Hence it is solvable and its solution is x = 2 and y = 4

p – q = 3 q = p - 3 ... equation (i)

Equation (i) will be plotted as line AB.

2p – 3q = 36 q = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph, we can see that two lines are intersecting each other at point P(--27,-30). Hence it is solvable and its solution is x = -27 and y = -30

p – q = 3 q = p - 3 ... equation (i)

Equation (i) will be plotted as line AB.

p – q = 15 q = p - 15 ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph, we can see that these two lines are parallel to each other. Hence it is not solvable.

p – q = 3 q = p - 3 ... equation (i)

Equation (i) will be plotted as line AB.

8p – 8q = 5 q = ... equation (ii)

Equation (ii) will be plotted as line CD.

Here in the graph, we can see that these two lines are parallel to each other. Hence it is not solvable.

x + y = 5 …(1)

Let us express the equations (1) in the form of

a₁x + b₁y + c₁ = 0 where a and b can’t be 0 at the same time.

Let the required equation be a₂x + b₂y + c₂ = 0

x + y = 5

∴ x + y + (-5) = 0

Or 1 × x + 1 × y + (-5) = 0

Here a₁ = 1, b₁ = 1, c₁ = -5

For lines of equation to be parallel in the graph, we need

Let and where k₁ and k₂ are any real number and k₁≠k₂ also both are non zero.

For simplicity, take k₁ = 0.5 and k₂ = 0.2

So, ,

and

Therefore, required equation is 2x + 2y – 25 = 0 which will be parallel in the graph with equation x + y = 5

x + y = 5 …(1)

Let us express the equations (1) in the form of

a₁x + b₁y + c₁ = 0 where a and b can’t be 0 at the same time.

Let the required equation be a₂x + b₂y + c₂ = 0

x + y = 5

∴ x + y + (-5) = 0

Or 1 × x + 1 × y + (-5) = 0

Here a₁ = 1, b₁ = 1, c₁ = -5

For lines of equation to be intersecting in the graph, we need

Let , and c₁ = c₂ where k₁ and k₂ are any real number

and k₁≠k₂ also both are non zero.

For simplicity, take k₁ = 0.5 and k₂ = 0.2

So, , and

Therefore, required equation is 2x + 5y – 5 = 0 which will be intersecting in the graph with equation x + y = 5

x + y = 5 …(1)

Let us express the equations (1) in the form of

a₁x + b₁y + c₁ = 0 where a and b can’t be 0 at the same time.

Let the required equation be a₂x + b₂y + c₂ = 0

x + y = 5

∴ x + y + (-5) = 0

Or 1 × x + 1 × y + (-5) = 0

Here a₁ = 1, b₁ = 1, c₁ = -5

For lines of equation to be parallel in the graph, we need

Let where k is any real number and k≠0.

For simplicity, take k= 0.5

So, ,

and

Therefore, required equation is 2x + 2y – 10 = 0 which will be overlapping in the graph with equation x + y = 5

Rewriting the above equations in the standard form of ax + by = c

8x + 5y = 11 (i)

3x – 4y = 10 (ii)

In this problem, we will eliminate ‘y’ from both the equations (i) & (ii),

Note: - We can eliminate any one of the variable, to eliminate y, we have to make the coefficient of y in both the equations equal.

We have to multiply the coefficient of y with such a number that makes them equal to the common factor of both of them.

Multiply equations (i) with 4 and (ii) with 5 .

⇒ x = 2, putting the value of x = 2 in equation (i), we get y = - 1

Hence the solution of the two linear equation is x = 2 and y = - 1

Let us take the equation (i) as Straight Line A and (ii) as B . To draw the lines we need at least 2 points to draw.

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

For 8x + 5y = 11

For 3x – 4y = 10

Hence also by graphical method the solution comes out to be (2, - 1)

Rewriting the above equations in the standard form of ax + by = c

2x + 3y = 7 (i)

3x + 2y = 8 (ii)

In this problem, we will eliminate ‘x’ from both the equations (i) & (ii)

Multiply equations (i) with 3 and (ii) with 2 to make the coefficient of x equal.

⇒ y = 1, Putting the value of y in equation (i) gives x = 2

Hence the solution of this linear equation is x = 2 and y = 1.

Let us take the equation (i) as Straight Line A and (ii) as B . To draw the lines we need at least 2 points to draw

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight\ line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

For 2x + 3y = 7

For 3x + 2y = 8

Hence, Also by graphical method the solution comes out to be (2, 1).

To eliminate y, we have to make the coefficient of y in both the equations equal, so we have to multiply the coefficient of y with such a number that makes them equal to the common multiple of both of them.

In this case, equation 7x – 5y + 2 = 0 should be multiplied by 3 such that the coefficient of y will be equal to 15 in both the cases.

In this case, the coefficient of x are 4 and 6, thinking of the common multiple of the 4 and 6 comes out to be 12, hence multiplying 4x – 3y = 16 by 3 and 6x + 5y = 62 by 2 makes the coefficient of x as 12 and then subtracting both the equations to eliminate x.

In this problem, we will eliminate ‘y’ from both the equations

3x + 2y = 6 (i)

2x – 3y = 17 (ii)

Multiplying equation (i) with 3 and (ii) with 2, We have

⇒ x = 4

Putting the value of x in equation (i)

We have y = 1/2 (6 - 3x)

y = - 3.

Hence the solution of the linear equation is x = 4 and y = - 3.

Rewriting the above equations in standard form,

2x + 3y = 32 (i)

- 9x + 11y = 3 (ii)

In this problem, we will eliminate ‘x’ from both the equations.

Hence multiplying equation (i) with 9 and (ii) with 2, We have

⇒ y = 6

Putting the value of y in equation (i)

we have, x = 1/2 (32 - 3y)

⇒ x = 7

Hence the solution of the linear equation is x = 7 and y = 6.

Rewriting the above equations in the standard form of ax + by = c.

x + y = 48 (i)

x – 2.5 y = 6 (ii)

In this problem, we will eliminate ‘x’ from both the equations, simply subtracting

⇒ y = 12

Putting the value of y in equation(i)

we have, x = (48 - y) or x = 36.

Hence the solution of the linear equation is x = 36 and y = 12.

Rewriting the above equation as.

3x + 2y = 48 (i)

5x – 12y = - 12 (ii)

In this problem, we will eliminate ‘y’ from both the equations!

Multiplying equation (i) with 6, we have

⇒ x = 12

Putting the value of y in equation(i)

We have y = 1/2(48 - 3x) or y = 6.

Hence the solution of the linear equation is x = 12 and y = 6.

Let as Y, Rewriting the above equation as.

3x – 2Y = 5 (i)

x + 4Y = 4 (ii)

In this problem, we will eliminate ‘y’ from both the equations. Multiplying equation (i) with 2, we have

⇒ x = 2

Putting the value of y in equation(i)

We have Y = - 1/2 (5 - 3x) or Y = 1/2

Since, , therefore y = 2.

Hence the solution of the linear equation is x = 2 and y = 2

Rewriting the above equation as.

3x + 2y = 6 (i)

2x + 3y = 6 (ii)

In this problem, we will eliminate ‘y’ from both the equations,

Multiplying equation (i) with 3 and (ii) with 2, we have

⇒

Putting the value of x in equation(i)

We have y = 1/2(6 - 3x) or

Hence the solution of the linear equation is and

Rewriting the above equation as.

5x – 3y = 8 (i)

9x + 7y = 126 (ii)

In this problem, we will eliminate ‘y’ from both the equations. Multiplying equation (i) with 7 and (ii) with 3, we have

⇒ x = 7, putting the value of x in equation(i),

we have or y = 9.

Hence the solution of the linear equation is x = 7 and y = 9.

Rewriting the above equation as.

x + y = 5xy (i)

x - y = 9xy (ii)

In this problem, we will eliminate ‘y’ from both the equations.

Adding both the equations

2x = 14xy (cancelling x from both the sides)

⇒

Putting the value of y in equation(i)

x = 5xy - y or x = - 1/2

Hence the solution of the linear equation is x = and y = -1/2

Note: - Since we have cancelled x as it is common from both the sides, we have neglected a solution that is x = 0, similarly other solution as y = 0.

Let us take as

and

Then the above equations will be

1a + 1.b = 3 (i)

2a + 3b = 5 (ii)

In this problem, we will eliminate ‘a’ from both the equations, Multiplying equation (i) with 2, we have

⇒ b = - 1, putting the value of b in equation(i) a = 3 - b or a = 4.

Since ⇒ and

⇒ y = 1

Hence the solution of the linear equation is and y = 1.

Let us take as and

Then above equations can be written as

14a + 3b = 5 (i)

21a - b = 2 (ii)

In this problem, we will eliminate ‘b’ from both the equations.

Multiplying equation (ii) with 3, we have

⇒

Putting the value of a in equation(i) gives b = 1

Now Since and It will form again a system of linear equations in x and y.

Rearranging the above equation

x + y = 7 (iii)

x – y = 1 (iv)

⇒ x = 4, putting the value of x in equation(iii) gives y = 3.

Hence the solution of the linear equation is x = and y = 3.

Rewriting the above equation as.

9y - x = 7 (i)

5y - x = - 5 (ii)

In this problem, we will eliminate ‘x’ from both the equations by subtracting the equation.

y = 3, putting the value of y in equation(i), we have x = 20

Hence the solution of the linear equation is y = 3 and x = 20

In this problem, we will eliminate ‘y’ from both the equations by subtracting the equation 2nd from 1st.

x + y = a + b (i)

(ii)

Multiplying equation 1st by ‘b’ to make the coefficient of y equal in both equation. So we get,

⇒

Putting the value of x in equation (i) we get, value of y as

⇒

⇒

⇒

Hence the solution of the linear equation is and .

Rewriting the above equation as

…(i)

…(ii)

Multiplying equation (i) by ‘b’ and equation (ii) by ‘a’ to eliminate y, we get

⇒ x = 1

Putting the value of x in (i) we get

Hence the solution of the linear equation is

In this case, we will eliminate y from both the equations

ax + by = c (i)

(ii)

Multiplying equation (i) by ‘b’ we get,

Putting the value of x in equation (i) we get,

Hence the solution of the linear equation is and

In this case, we will eliminate y from both the equations

(i)

(ii)

Multiplying equation (i) by ‘a’ and (ii) by ‘b’ to eliminate y we get,

a²x + aby = a (iii)

(iv)

Subtracting (iii) from (iv), we get

Putting the value of x in (i) we get

Hence the solution of the linear equation is

This equation is of the form

This is possible if both ‘a’ = 0 and b = 0

So, for above equation equals to zero

and

Now multiplying equation (i) by 2 and adding both equation we get,

⇒ x = 1

Putting the value of x in (i) we get

So for the value of x = 1 and y = 1 the given equation equals to zero

Taking L.C.M

⇒ 2x + 3y = 48

Taking L.C.M

⇒ 2y + 7x = xy

⇒ 7x = xy - 2y

⇒ 7x = y(x - 2)

Let’s express the variable in terms of another variable.

⇒ 2(x - y) = 3 ………[1]

⇒ 2x - 2y = 3

⇒ 2x = 3 + 2y

………[3]

and,

5x + 8y = 14 ………[2]

⇒ 5x = 14 – 8y

………[4]

on comparing eq. [3] and eq. [4]

we get,

⇒ 15 + 10y = 28 – 16y

⇒ 26y = 13

………[5]

from eq. [3]

………[6]

by eq. [5] and [6]

x = 2,

by using the result in eq. [1] and eq. [2]

2(x - y) = 3

L.H.S

= 3 = R.H.S

similarly,

by eq. [2]

5x + 8y = 14

L.H.S

= 10 + 4

= 14 = R.H.S

expressing the equations in terms of one variable.

………[1]

………[2]

⇒

⇒ 2xy + 3 = 5y

⇒ 2xy = 5y – 3

………[3]

by eq. 2

⇒ 5xy – 2 = 3y

⇒ 5xy = 3y + 2

………[4]

on comparing eq. [3] and eq. [4]

⇒ 5y(5y – 3) = 2y(3y + 2)

⇒ 25y² – 15y = 6y² + 4y

cancelling one power of y from both sides

⇒ 25y – 15 = 6y + 4

⇒ 19y = 19

⇒ y = 1

by eq. [3]

⇒ x = 1

∴ solution of this pair is x = 1, y = 1

verifying by eq. [1]

L.H.S

= 5 = R.H.S

by eq. [2]

L.H.S

= 3 = R.H.S

………[1]

………[2]

by eq. [1]

taking L.C.M

⇒ 3x + 2y = 6

………[3]

by eq. [2]

⇒ 2x + 3y = 6

………[4]

on comparing eq. [3] and eq. [4]

⇒ 12 – 4y = 18 – 9y

⇒ 5y = 6

………[5]

by using eq. [5] in eq. [3]

………[6]

by eq. [5] and [6]

verifying the result by eq. [1]

L.H.S

verifying the result by eq. [2]

L.H.S

4x – 3y = 18 ………[1]

4y – 5x = - 7 ………[2]

by eq. [1]

4x – 3y = 18

⇒ 4x = 18 + 3y

………[3]

by eq. [2]

4y – 5x = - 7

⇒ 5x = 4y + 7

………[4]

on comparing eq. [3] and eq. [4]

⇒ 90 + 15y = 28 + 16y

⇒ y = 62 ………[5]

by using eq. [5] in eq. [3]

………[6]

by eq. [5] and [6]

,

verifying the result by eq. [1]

L.H.S

verifying the result by eq. [2]

L.H.S

expressing the equations in terms of one variable.

2x + y = 8

………[1]

2y – 3x = –5

⇒ 3x = 2y + 5

………[2]

by comparing eq. [1] and eq. [2]

⇒24 – 3y = 4y + 10

⇒7y = 14

⇒ y = 2 ………[3]

by using [3] in [1]

Let’s make the graphs of given linear equations,

We get the following tables for the given linear equations.

For 2x + y = 8

y = 8 – 2x

For 2y - 3x = - 5

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines intersect at (3, 2), hence the solution to this pair is (3, 2).

expressing the equations in terms of one variable

3x – 2y = 2

………[1]

7x + 3y = 43

………[2]

by comparing eq. [1] and eq. [2]

⇒ 14y + 14 = 129 – 9y

⇒ 23y = 115

⇒ y = 5 ………[3]

by using [3] in [1]

⇒ x = 4

expressing the equations in terms of one variable

2x – 3y = 8

………[1]

⇒ 3x + 3y = 7x – 7y

⇒ 4x = 10y

………[2]

by comparing eq. [1] and eq. [2]

⇒ 8 + 3y = 5y

⇒ 2y = 8

⇒ y = 4 ………[3]

by using [3] in [1]

⇒ x = 10

expressing the equations in terms of one variable

⇒ 4x – 4y = 3y – 3

⇒ 4x = 7y – 3

………[1]

⇒ 4x – 5y = 7x – 49

⇒ 3x = 49 – 5y

………[2]

by comparing eq. [1] and eq. [2]

⇒ 21y – 9 = 196 – 20y

⇒ 41y = 205

⇒ y = 5 ………[3]

by using [3] in [1]

expressing the equations in terms of one variable

⇒ 5x + 5 = 4y + 4

………[1]

⇒ 2x – 10 = y – 5

⇒ 2x = 5 + y

………[2]

by comparing eq. [1] and eq. [2]

⇒ 8y – 2 = 25 + 5y

⇒ 3y = 27

⇒ y = 9 ………[3]

by using [3] in [1]

⇒ x = 7

expressing the equations in terms of one variable

x + y = 11

………[1]

⇒ 8y + 16 = 10y + x

………[2]

by comparing eq. [1] and eq. [2]

⇒ 16 – 2y = 11 – y

⇒ y = 5 ………[3]

by using [3] in [1]

⇒ x = 11 – y

⇒ x = 11 - 5

⇒ x = 6

expressing the equations in terms of one variable

⇒ 4x + 3y = 12

………[1]

2x + 4y = 11

⇒ 2x = 11 – 4y

………[2]

by comparing eq. [1] and eq. [2]

⇒ 24 – 6y = 44 – 16y

⇒ 10y = 20

⇒ y = 2 ………[3]

by using [3] in [1]

expressing the equations in terms of one variable

⇒ xy + 2 = 7y

………[1]

⇒ 2xy – 6 = 9y

………[2]

by comparing eq. [1] and eq. [2]

⇒ 14y – 4 = 9y + 6

⇒ 5y = 10

⇒ y = 2 ………[3]

by using [3] in [1]

⇒ x = 6

expressing the equations in terms of one variable

⇒ 6y + 6x = 5xy

⇒ x(5y - 6) = 6y

………[1]

⇒ 6y – 6x = xy

⇒ x(y + 6) = 6y

………[2]

by comparing eq. [1] and eq. [2]

⇒ y + 6 = 5y – 6

⇒ 4y = 12

⇒ y = 3 ………[3]

by using [3] in [1]

expressing the equations in terms of one variable

⇒ x + y = 2xy

⇒ x(2y - 1) = y

………[1]

⇒ x – y = xy

⇒ x(1 – y) = y

………[2]

by comparing eq. [1] and eq. [2]

⇒ 2y – 1 = 1 – y

⇒ 3y = 2

………[3]

by using [3] in [1]

expressing the equations in terms of one variable

⇒ 4x + 4y + 5x – 5y = 20× 5

⇒ 9x – y = 100

………[1]

⇒ 5x + 5y + 4x – 4y = 116

⇒ 9x = 116 - y

………[2]

by comparing eq. [1] and eq. [2]

⇒ 2y = 16

⇒ y = 8 ………[3]

by using [3] in [1]

expressing the equations in terms of one variable

⇒ 8 – xy = - 2x

⇒ x(y – 2) = 8

………[1]

⇒ 8 + 2xy = 10x

⇒ x(10 – 2y) = 8

………[2]

by comparing eq. [1] and eq. [2]

⇒ 10 – 2y = y – 2

⇒ 3y = 12

⇒ y = 4 ………[3]

by using [3] in [1]

⇒ x = 4

expressing the equations in terms of one variable

2 – 2(3x – y) = 10

⇒ 2 – 6x + 2y = 10

⇒ 6x = 2y – 8

dividing by 3 on both sides.

⇒ 3x = y – 4

………[1]

10(4 – y) – 5x = 4(y – x)

⇒ 40 – 10y – 5x = 4y – 4x

⇒ x + 14y = 40

⇒ x = 40 – 14y ………[2]

by comparing eq. [1] and eq. [2]

⇒ y – 4 = 120 – 42y

………[3]

by using [3] in [1]

We have to express x of the equation in terms of y.

∴ x of the given equation in terms of variable y is .

Given let be put instead of y in the equation 2x + 3y = 9.

We have to find the value of x.

⇒ 22x – 21 = 9 (5)

⇒ 22x – 21 = 45

⇒ 22x = 45 + 21

⇒ 22x = 66

⇒ x = 3

∴ Value of x is 3.

Given 3x – y = 7 … (1)

2x + 4y = 0 … (2)

Expressing y of equation (1) in terms of x,

⇒ 3x – y = 7

⇒ 3x - 7 = y … (3)

Substituting (3) in (2),

⇒ 2x + 4y = 0

⇒ 2x + 4 (3x – 7) = 0

⇒ 2x + 12x – 28 = 0

⇒ 14x – 28 = 0

⇒ 14x = 28

∴ x = 2

Substituting x value in (3),

⇒ 3x – 7 = y

⇒ 3 (2) – 7 = y

∴ y = -1

∴ By solving, we get x = 2 and y = -1.

Justification:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

For 3x – y = 7

⇒ y = 3x – 7

For 2x + 4y = 0

⇒ x = -2y

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines intersect at (2, -1), hence the solution to this pair is (2, -1).

Hence x and y values satisfy equations (1) and (2).

Given … (1)

And … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 12 + 4y = 2 (12)

⇒ 12 + 4y = 24

⇒ 4y = 24 – 12

∴ y = 3

Substituting y value in (3),

∴ x = 2

∴ By solving, we get x = 2 and y = 3.

Justification:

Generally, we substitute x = 0 or y = 0 in the given linear equations to get y and x. So we get two points on the straight line. To find more points on the line, take different values of x related to it, we get different values for y from the equation.

We get the following tables for the given linear equations.

For

For

∴ x = 8 – 2y

Plotting these points on a graph and joining them, we get two straight lines.

From the graph, we can see that both lines intersect at (2, 3), hence the solution to this pair is (2, 3).

Hence we can say that x and y values satisfy equations (1) and (2).

… (1)

And … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 12 (5y – 19) = 11 (2y)

⇒ 60y – 228 = 22y

⇒ 60y – 22y = 228

⇒ 38y = 228

⇒ y = 228/ 38

∴ y = 6

Substituting y value in (3),

∴ x = 1/4

∴ By solving, we get x = 1/4 and y = 6.

Justification:

(1) -

⇒ 1 = 1

And (2) -

∴ We can see x = 1/4 and y = 6 satisfy equations (1) and (2).

Given … (1)

And

… (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 6 (10y + 5) = 35 (2y)

⇒ 60y + 30 = 70y

⇒ 30 = 70y – 60y

⇒ 30 = 10y

⇒ y = 30/ 10

∴ y = 3

Substituting y value in (3),

∴ x = 2

∴ By solving, we get x = 2 and y = 3.

Justification:

(1) -

⇒ 1 + 1 = 2

⇒ 2 = 2

And (2) -

∴ We can see that x = 2 and y = 3 satisfy equations (1) and (2).

Given … (1)

And … (2)

Expressing x of equation (1) in terms of y,

⇒ x + y = 3xy

∴ x = 3xy – y … (3)

Substituting (3) in (2),

⇒ 3xy – 2y = xy

⇒ 3xy – xy = 2y

⇒ 2xy = 2y

∴ x = 1

Substituting x value in (3),

⇒ x = 3xy – y

⇒ 1 = 3(1) y – y

⇒ 1 = 3y – y

⇒ 1 = 2y

∴ y = 1/2

∴ By solving, x = 1 and y = 1/2.

Justification:

(1) -

⇒ 3 = 3

And (2) -

⇒ 1 = 1

∴ We can see that x = 1 and y = 1/2 satisfy equations (1) and (2).

Given … (1)

And … (2)

Expressing x of equation (2) in terms of y,

… (3)

Substituting (3) in (1),

Substituting y value in (3),

∴ x = 1/2

∴ By solving, we get x = 1/2 and

Justification:

(1) -

And (2) -

∴ We can see that x = 1/2 and satisfy the equations (1) and (2).

Given 2 (x – y) = 3

⇒ 2x – 2y = 3 … (1)

And 5x + 8y = 14 … (2)

Expressing x of (1) in terms of y,

⇒ 2x – 2y = 3

⇒ 2x = 3 + 2y

… (3)

Substituting (3) in (2),

⇒ 5x + 8y = 14

⇒ 15 + 26y = 28

⇒ 26y = 28 – 15

⇒ 26y = 13

⇒ y = 13/ 26

∴ y = 1/2

Substituting y value in (3),

∴ x = 2

∴ By solving, we get x = 2 and y = 1/2.

Given … (1)

And … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 21y – 15 = 6y

⇒ 21y – 6y = 15

⇒ 15y = 15

∴ y = 1

Substituting y value in (3),

∴ x = 1

∴ By solving, we get x = 1 and y = 1.

Given … (1)

And … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 12 + 5y = 18

⇒ 5y = 18 – 12

⇒ 5y = 6

∴ y =

Substituting y value in (3),

∴ By solving, we get and .

Given … (1)

And 7x – 5y = 2 … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 7x – 5y = 2

∴ y = 8

Substituting y value in (3),

∴ x = 6

∴ By solving, we get x = 6 and y = 8.

Given … (1)

And … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 20 (3y – 11) = 19 (2y)

⇒ 60y - 220 = 38y

⇒ 220 = 60y – 38y

⇒ 220 = 22y

⇒ y = 220/ 22

∴ y = 10

Substituting y value in (3),

∴ x = 4

∴ By solving, we get x = 4 and y = 10.

Given … (1)

And … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 4 (2y – 3) = 7 (7y – 31)

⇒ 8y – 12 = 49y - 217

⇒ 217 – 12 = 49y – 8y

⇒ 205 = 41y

∴ y = 5

Substituting y value in (3),

∴ x = 8

∴ By solving, we get x = 8 and y = 5.

Given … (1)

And … (2)

Expressing x of equation (1) in terms of y,

… (3)

Substituting (3) in (2),

⇒ 630 – 62y = 72

⇒ 630 – 72 = 62y

⇒ 558 = 62y

∴ y = 9

Substituting y value in (3),

∴ x = 7

∴ By solving, we get x = 7 and y = 9.

p (x + y) = q (x – y) = 2pq

⇒ px + py = 2pq … (1)

⇒ qx – qy = 2pq … (2)

Expressing x of equation (1) in terms of y,

⇒ px + py = 2pq

⇒ px = 2pq – py

⇒ x = 2q – y … (3)

Substituting (3) in (2),

⇒ q (2q – y) – qy = 2pq

⇒ q (2q – y – y) = 2pq

⇒ 2q – 2y = 2p

⇒ 2 (q – y) = 2p

⇒ q – y = p

∴ y = p – q

Substituting y value in (3),

⇒ x = 2q – y

⇒ x = 2q – (p – q)

∴ x = 3q – p

∴ By solving, we get x = 3q – p and y = p – q.

8x + 5y - 1 = 0

3x - 4y - 10 = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

and

3x - 4y - 1 = 0

4x - 3y - 6 = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 3 and y = 2

5x + 3y - 11 = 0

2x - 7y + 12 = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 1 and y = 2

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 4 and y = - 1

⇒ x - 2y = 24 (Multiplying both sides by 6)

⇒ x - 2y - 24 = 0

⇒ x - 8y - 48 = 0 (Multiplying both sides by 12)

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 16 and y = - 4

⇒ 3x + 5y = 0 (Multiplying both sides by 15)

⇒ 15x - 20y - 9 = 0 (Multiplying both sides by 60)

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 1/3 and y = ( - 1/5)

⇒ 4(x + 2) + 7(y - x) = 28(2x - 8) (Multiplying both sides by 28)

⇒ 59x - 7y - 232 = 0

⇒ (2y - 3x) + 6y = 3(3x + 4) (Multiplying both sides by 3)

⇒ 12x - 8y + 12 = 0

⇒ 3x - 2y + 3 = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 5 and y = 9

x + 5y - 36 = 0

⇒ 3(x + y) = 5(x - y) (Cross Multiplication)

⇒ 2x - 8y = 0

⇒ x - 4y = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 16 and y = 4

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = 21 and y = 24

x + y - 2b = 0

x - y - 2a = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = (a + b) and y = (b - a)

x - y - 2a = 0

ax + by - (a² + b²) = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = (a + b) and y = (b - a)

⇒ bx + ay = 2ab (Multiplying both sides by ab)

⇒ bx + ay - 2ab = 0

ax - by - (a² - b²) = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒ x = a and y = b

ax + by - 1 = 0

bx + ay - = 0

To solve the equation by cross multiplication method, we use:

Putting the values, we get:

So, and

⇒