# Free Pair of Linear Equations in Two Variables 01 Practice Test - 10th Grade

### Question 1

If (2,2) lies on 4x+5y=k, the value of k is

#### SOLUTION

Solution :If the point (2,2) lies on the given linear equation, then it must satisfy the equation.

Putting the value of x and y in the equation we get:

⇒4(2)+5(2)=k

⇒k=18

Thus, the value of k is 18.

### Question 2

Which of the following equations have x = 7, y = 3 as solution?

x + y = 4

x + 3y = 16

x - y + 4 = 0

x + y = 7

#### SOLUTION

Solution :B

The solution of the linear equation should satisfy the equation. The given values of x and y are 7 and 3 respectively.

Among the given equations, only the equation x + 3y = 16 is able to satisfy the equation.

⇒7+3(3)=16

Hence x=7 and y=3 is the solution of x + 3y = 16.

### Question 3

Half the perimeter of a rectangular room is 46 m, and its length is 6 m more than its breadth. What is the length and breadth of the room?

2 m, 20 m

26 m, 20 m

56 m, 40 m

2 m, 3 m

#### SOLUTION

Solution :B

Let l and b be the length and breadth of the room respectively.

Then, the perimeter of the room is 2(l+b) metres.

From the question,

l=6+b ...(1)

12×2(l+b)=46

⇒l+b=46 ...(2)

Let's solve these two equations using substitution method.

On substituting the value of l from (1) in (2), we get

6+b+b=46

⇒ 6+2b=46

⇒2b=40

⇒b=20 m

⇒l=20+6=26 m

Therefore, length and breadth are 26 m and 20 m long respectively.

### Question 4

Which of the following is a solution to 3x+4y=38?

(3, 4)

(6, 5)

(2, 19)

(3, 12)

#### SOLUTION

Solution :B

The solution of a linear equation satisfies the equation.

3×3+4×4=9+16=25≠38→ (3, 4) does not satisfy the equation.

3×6+4×5=18+20=38→ (6, 5) satisfies the equation.

3×2+4×19=6+76=82≠38→ (2, 19) does not satisfy the equation.

3×3+4×12=9+48=57≠38→ (3, 12) does not satisfy the equation.

### Question 5

______ satisfies the equation 3x+4y=2.

(2, 1)

(-2, 1)

(2, -1)

(2, 2)

#### SOLUTION

Solution :C

The solutions of a linear equation must be able to satisfy the equation. Among the given options, point (2,-1) satisfies the given equation. Substituing the value of x and y in the equation we get:

3(2) + 4(-1) = 2.

### Question 6

On the basis of the graph shown below which shows the graphical representation of a pair of linear equations**, **the pair of linear equations has _______________solution(s).

a unique

infinite

four

zero

#### SOLUTION

Solution :A

If the graph of linear equations represented by the lines intersect at only one point, then the point is its only solution.

Here, the lines meet at only one point (1,-1). Therefore, the given pair of equations has a unique solution.

### Question 7

Solve the following system of equations:

8v−3u=5uv

6v−5u=−2uv

#### SOLUTION

Solution :B

Divide the given equations by uv,

8v−3u=5uv⇒8u−3v=5...(1)

6v−5u=−2uv⇒6u−5v=−2...(2)

Assume 1u=x and 1v=y

Put the values of 1u and 1v in (1) and (2)

8x−3y=5...(3)

6x−5y=−2...(4)

Solving equations (3) and (4) we get x and y

Multiply (3) with 5 and (4) with 3 to equate the coefficients of y.

40x−15y=25

18x−15y=−6

⇒x=3122⇒u=2231

Substituting x in (3)

8×3122−3y=5→3y=6911

→y=2311⇒v=1123

So, u=2231 and v=1123

### Question 8

If y=12(3x+7) is written in the form ax+by+c=0, which of the following are possible values of a, b and c?

a=+3, b=−2, c=+7

a=+3, b=+2, c=+7

a=−3, b=+2, c=−7

a=−3, b=−2, c=−7

#### SOLUTION

Solution :A and C

The given equation is y=12(3x+7).

Simplifying the equation we get

2y−3x−7=0

Rewriting the equation in standard form ax + by + c = 0 we get

−3x+2y−7=0

∴ On comparing the above equation with standard form ax + by +c = 0, we get the values of a, b and c is -3, 2 and -7 respectively.

But the equation can also be written as,

3x−2y+7=0

(By multiplying both the sides by (-1))

∴ On comparing the above equation with standard form ax + by +c = 0, we get the value of a, b and c is +3, -2 and +7 respectively.

### Question 9

The solution of the pair of linear equations

x+2y=5 and 7x+3y=13 is (2, 1).

True

False

#### SOLUTION

Solution :B

x+2y=5 ...(1)

7x+3y=13 ...(2)

Let's solve these equations using elimination method.

On multiplying the first equation by 7, we get7x+14y=35 ...(3)

On subtracting (2) from (3), we have

7x+14y=35

−7x−3y=−13

_______________11y=22

⇒y=2On subsituting value of y in x+2y=5, we get

x+2(2)=5

⇒x=1⇒x=1 and y=2

Hence, the solution of the given equations is (1, 2).

### Question 10

Solve the equations for x and y.

2x−3y=7

5x+y=9

3,4

2, -1

2,2

3, -1

#### SOLUTION

Solution :B

Given

2x−3y=7...(i)

5x+y=9...(ii)

Rearranging (ii), we get y=9−5x...(iii)

Substituting (iii) in (i), we get

2x−3(9−5x)=7

⇒17x=34

⇒ x=2.

Substituting the value of x in (i), we get

2(2)−3y=7

⇒ y=−1.