Linear Equation In Two Variables

In the above linear equation in 2 variables, expressing y in terms of x we get,

– 5y = 7 – 2x

for any value of x we get a different value of y.

Hence, C is the correct option.

A. on (1, 1) is a solution to the given equation

B. If positive real numbers are chosen, it will have many solutions.

C. If real numbers are chosen, it will have infinite solutions.

D. If rational numbers are chosen, it will have many solutions.

Hence A is the correct option.

As (2, 0) = (x, y)

putting values of x and y in the above equation, we get

2×2 + 3×0 = k

k = 4

Hence A is the correct answer.

Solving the above equation we get,

As the coefficient of y is 0, therefore, y can take any value and will not affect our answer.

A.

y = any value

B.

C.

D.

Hence A is the correct option.

let cut the y-axis at P. therefore at P x-coordinate = 0.

putting x = 0, we get

Hence the coordinates are (0, 2).

A. is wrong because it has x = 2

B. is wrong because it has y = 3

C. is wrong because it has x = 3

D. is right because it has x = 0 and y = 2

A. Simplifying the equation we get x + y = 7

B. Simplifying the equation we get x + 0y = 7 which is equal to x = 7

C. Simplifying the equation we get y = 7

D. simplifying the equation we get 0x + 0y = 7 which is not possible.

Hence B is the correct answer.

A point on the x-axis will have its y- coordinates as 0 and any value as x.

TIP: Any point on x-axis will have y = 0 and any point on y-axis will have x = 0.

A. it has a y coordinate.

B. As x = 0 it is a point on y axis.

C. y = 0 and x has a value.

D. Both x and y are equal to x.

Therefore C is the correct option.

In the equation we have x = y

(x, y) = (a, a)

A. Both x and y are equal

B. Both x and y are not equal

C. Both x and y are not equal

D. Both x and y are not equal

Hence A is the correct option.

TIP: Equation of a line parallel to x-axis is y = a.

and equation of a line parallel to y-axis is x = a.

Equation of x-axis is y = a.

As at x-axis,

we get equation of x-axis as

Hence B is the correct answer.

From the TIP given in previous question we get that the line is parallel to x-axis.

As y = 6

we get that it has y coordinate of 6.

Hence A is the correct answer.

Putting values in the options

A. 5 + 4 = 7

not possible

B. 25 + 4 = 7

not possible

C. 5 + 2 = 7

D. 25 + 2 = 7

not possible

Hence C is the correct answer.

In the above points given (x, y) = (a, – a)

i.e. (2, – 2); (0, 0); ( – 2, 2).

Therefore, x = – y

x + y = 0

simplifying above options

A. x = y

B. x = – y

C.

D.

Hence B is the correct option.

Positive solutions of the equation will give us positive values for x and y.

As x and y both are positive it must lie in the 1st quadrant.

A. IN 1st quadrant both x and y are positive.

B. In the 2nd quadrant x is negative and y is positive.

C. in the 3rd quadrant both x and y are negative.

D. In the 4th quadrant x is positive and y is negative.

Hence A is the correct Answer.

At the x-axis, y coordinate = 0

putting y = 0 in the above equation.

2x = 6

x = 3

Therefore, (x, y) = (3, 0)

Hence the correct answer is C.

As y = x

x coordinate = y coordinate.

A.

B.

C. 1 = 1

D.

Hence C is the correct option.

Consider an equation x = y

if it is multiplied by a non-zero number, it gets cancelled. Therefore the linear equations remains the same.

ax = ay

Hence A is the correct option.

Through any one point Infinite number of lines can be passed.

Hence through the point (x, y) = (1, 2)

infinite number of lines and hence infinite linear equations can be satisfied.

Hence C is the correct option.

The point(a, a) has both its x and y coordinates = a.

Therefore we get, x = y = a

A. Point on x-axis is of the form (x, 0)

B. Point on y-axis is of the form (0, y)

C. Point on line x = y is of the form (a, a)

D. Point on line x + y = 0 is of the form (a, – a)

Hence C is the correct option.

In the point (a, – a) we have x coordinate as a , y coordinate as ( – a)

Hence we get

Taking y to the left hand side, we get

x + y = 0

A. Point on line x = a is of the form (a, y)

B. Point on line y = a is of the form (x, a)

C. Point on line x = y is of the form (a, a)

D. Point on line x + y = 0 is of the form (a, – a)

Hence D is the correct option.

True.

Explanation:

Given equation is 3x + 4y = 12.

Putting the values of solution x = 0 and y = 3 in the equation

3(0) + 4(3) = 12 = RHS.

False.

Explanation:

Putting the values of solution x = 0 and y = 7 in the equation

0 + 2(7) = 14 ≠RHS

True.

Explanation:

Given equation: x + y = 0

∴x = – y

Given solution: ( – 3, 3) and ( – 1, 1) which satisfy the given equation.

True

Explanation:

Given equation can be represented as 1.x + 0.y = 3

∴ For all values of y, x = 3 is true.

True.

Explanation:

Given equation: x – y + 2 = 0

Putting values of solution (0, 2) in the equation

0 – 2 + 2 = 0 = RHS

Similarly, for other solution sets (1, 3), (2, 4), (4, 6), the equation is satisfied.

But, for (3, – 5) the equation is not satisfied.

3 – ( – 5) + 2 = 10≠RHS

Thus, the co-ordinates represent some of the solutions for given equation.

False

Explanation:

Graph is the line containing points that satisfy the given equation. Thus, each point on the line (graph) is a solution of the linear equation.

False

Explanation:

Linear equation in two variables is represented geometrically by a line whose points make up the collection of solutions of the equation.

The given linear equations are:

y = x ------- (i)

Following are some values x and y which satisfy the equation

y = -x ------- (ii)

Following are some values x and y which satisfy the equation (ii)

Plotting the above values in the same graph:

From the above graph it can be observed that the two lines y = x and y = – x intersect each other at (0, 0).

Given: 2x + 5y = 19…………(i)

ordinate is times its abscissa

⇒ y = x = x

Putting y = x in eq. (i)

We have

2x + 5 x = 19
.

Putting x = 2 in eq. (i)

We have

2x2 + 5y = 19

y = = 3.

Therefore point (2, 3) is the required solution.

The required linear equation is:

y = – 3

Therefore, y = – 3

Following are some values x and y which satisfy the equation

Plotting the above values:

The required linear equation is:

x + y = 10

Therefore, x = 10 – y

Following are some values x and y which satisfy the equation

Plotting the above values:

Given: linear equation such that each point on its graph has an ordinate(y) 3 times its abscissa(x).

⇒ y = 3x is the required linear equation.

Given: 3y = ax + 7

Since the point (3, 4) lies on the graph of 3y = ax + 7.

3(4) = a(3) + 7

Or, a = .

Therefore,

Solving the given equation 2x + 1 = x – 3,

We have 2x – x = – 3 – 1

Or, x = – 4

(i) The representation of the solution on the number line is as shown:

x = – 4 is treated as an equation in one variable.

(ii) Equation x = – 4 is same as

x + 0.y = – 4

which is a linear equation in the variables x and y.

This is represented by a line. Now all the values of y are

permissible because 0.y is always 0. However, x must satisfy the

equation x = – 4.

The permissible values of x and y can be tabulated as:

Plotting the above values:

Given: x + 2y = 8

(i) Equation of x-axis: y = 0

Putting y = 0 in equation x + 2y = 8

We have x + 2 (0) = 8

Or, x = 8.

Thus, the required point is (8, 0).

(ii.) Equation of y-axis: x = 0

Putting x = 0 in equation x + 2y = 8

We have 0 + 2y = 8;

Or, y = = 4.

Or, y = 4.

Thus, the required point is (0, 4).

Given: y = x is a solution to the equation 2x + cy = 8

Put x = y

We have: 2y + cy = 8

Or, cy = 8−2y

c = − 2 = − 2

Given:

y varies directly with x i.e. y is proportional to x.

⇒ y = kx, where k is some real number. ------------- (i)

Given y = 12 at x = 3

Putting x = 3 and y = 12 in eq. (i)

We have,

12 = 3k

⇒ k = = 4

Thus the given linear equation is y = 4x

At x = 5: y = 4 x 5 = 20.

The given equation is y = 9x – 7. To draw the graph of this equation, we need at least two points lying on the graph.

For x = 0, y = – 7, therefore, (0, – 7) lies on the graph.

For y = 0, , therefore, (0.8, 0) lies on the graph.

Now plot the points M(0, – 7) and N (0.8, 0) and join them to get the line and now extend this straight line to get more solutions.

Now, to check that the given points are lying on the graph either we can take x and y pt. and draw perpendicular to the x-axis and y-axis to check if they lie on the graph(here, all are lying(as shown)) or we can put the given points in the equations and see if they are satisfying the equation or not.

If the given point satisfies the equation that means points lie on the graph. Therefore we’ll put the value of x and find the value of y for every point using equation y = 9x – 7.

• For pt. A(1, 2), x = 1 and y = 2

∴ Putting x = 1 in y = 9x – 7, we get-

⇒ y = 9×1 – 7

⇒ y = 2, that means pt. A lies on the graph.

• For pt. B( – 1, – 16), x = – 1 and y = – 16

∴ Putting x = – 1 in y = 9x – 7, we get-

⇒ y = (9× – 1) – 7

⇒ y = – 16, that means pt. B also lies on the graph.

• For pt. C(0, – 7), x = 0 and y = – 7

∴ Putting x = 0 in y = 9x – 7, we get-

⇒ y = (9×0) – 7

⇒ y = – 7, that means pt. C also lies on the graph.

Conclusion: Point A, B, C lies on the graph of equation y = 9x – 7

The linear equation for the line will be-

y = mx + c

Where, c is the y-intercept, which from the graph is 2.

⇒ c = 2.

Taking x₁ = 6, y₁ = – 2 and x₂ = – 6, y₂ = 6

Now, m = slope of the line

∴ the linear equation will be-

⇒ 3y = – 2x + 6 (multiplying whole equation by 3)

⇒ 2x + 3y – 6 = 0

Now, the point where graph cuts:

(i)x-axis

For this, y = 0.

∴putting x = 0 in 2x + 3y – 6 = 0, we get-

⇒ 2x + 3×0 – 6 = 0

⇒ 2x = 6

⇒ x = 3

∴ the point where graph cuts x-axis is(3, 0).

(ii) y-axis

For this, x = 0.

∴putting x = 0 in 2x + 3y – 6 = 0, we get-

⇒ 2×0 + 3y – 6 = 0

⇒ 3y = 6

⇒ y = 2

∴ The point where graph cuts y-axis is (0, 2).

The given equation is 3x + 4y = 6. To draw the graph of this equation, we need at least two points lying on the graph.

Now, the point where graph cuts:

(i) X-axis

For this, y = 0.

∴putting x = 0 in 3x + 4y = 6, we get-

⇒ 3x + 4×0 = 6

⇒ 3x = 6

⇒ x = 2

∴ The point where graph cuts x-axis is (2, 0).

(ii) Y-axis

For this, x = 0.

∴ putting x = 0 in 3x + 4y = 6, we get-

⇒ 3×0 + 4y = 6

⇒ 4y = 6

⇒

∴ the point where graph cuts y-axis is(0, 1.5).

Now plot the points M (0, 1.5) and N (2, 0) and join them to get the straight line, i.e., graph of the equation.

(i)

°C

(ii)

350×9 = 5f – 160

5f = 3150 + 160

f = 662°F

(iii) If the temperature is 0°C, then the temperature in Fahrenheit will be-

∵

5f = 160

f = 32°F

And if the temperature is 0°F, then the temperature in Celsius will be-

∵

C = – 17.778

(iv) Since, both are equal, let C = f

⇒

⇒ 9f = 5f – 160

⇒ 4f = – 160

⇒ f = – 40°F

∴ the numerical value of the temperature which is same in both the scale is – 40

(i)

Put x = 313 in this equation, we get-

⇒ y = 72 + 32

⇒ y = 104°F

(ii)

Put y = 158 in the equation, we get-

790 = 9x – 2457 + 160 (multiplying whole equation by 5)

9x = 3087

x = 343K

Let the force be y, acceleration be x and constant mass be m.

Now, ∵ y α x (Newton’s second law of motion)

⇒ y = mx

Now, given that m = 6kg

⇒ y = 6x is the equation.

To draw the graph of this equation, we need at least three points lying on the graph.

For x = 0, y = 0, therefore, (0, 0) lies on the graph, i.e., graph passes through origin.

For x = 1, y = 6, therefore, (1, 6) lies on the graph.

For y = 1, , therefore, (0.17, 1) lies on the graph.

Now plot the points M (0, 0) and N (1, 6) and O () and join them to get the line and now extend this straight line to get more solutions.

Now, the values of y for given x can be calculated by putting the given values in equation y = 6x.

But it is given to do so by graph.

Now, to get the value of y from graph for given values of x, we’ll draw perpendicular from the point where a lies on the graph to the y-axis to get the value of corresponding y.

Therefore, by doing so, we get-

The force required when the acceleration produced is

(i) 5 ms ^{– 2} is 30N

(ii) 6 ms ^{– 2} is 36N

(where N(Newton) = kgms ^{– 2})