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

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.

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

A.

x + y = 4

B.

x + 3y = 16

C.

x - y + 4 = 0

D.

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.

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?

A.

2 m, 20 m

B.

26 m, 20 m

C.

56 m, 40 m

D.

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.

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

A.

(3, 4)

B.

(6, 5)

C.

(2, 19)

D.

(3, 12)

#### SOLUTION

Solution : B

The solution of a linear equation satisfies the equation.

3×3+4×4=9+16=2538 (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=8238 (2, 19) does not satisfy the equation.

3×3+4×12=9+48=5738 (3, 12) does not satisfy the equation.

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

A.

(2, 1)

B.

(-2, 1)

C.

(2, -1)

D.

(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.

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.

a unique

B.

infinite

C.

four

D.

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.

Solve the following system of equations:
8v3u=5uv
6v5u=2uv

A. u=0,v=0
B. u=2231,v=1123
C. Both A and B
D. u=3122, v= 2311

#### SOLUTION

Solution : B

Divide the given equations by uv,
8v3u=5uv8u3v=5...(1)
6v5u=2uv6u5v=2...(2)

Assume 1u=x and 1v=y
Put the values of 1u and 1v in (1) and (2)
8x3y=5...(3)
6x5y=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.
40x15y=25
18x15y=6 x=3122u=2231

Substituting x in (3)
8×31223y=53y=6911

y=2311v=1123

So, u=2231 and v=1123

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.

a=+3, b=2, c=+7

B.

a=+3, b=+2, c=+7

C.

a=3, b=+2, c=7

D.

a=3, b=2, c=7

#### SOLUTION

Solution : A and C

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

Simplifying the equation we get
2y3x7=0

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

3x+2y7=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,

3x2y+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.

The solution of the pair of linear equations
x+2y=5 and 7x+3y=13 is (2, 1).

A.

True

B.

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 get

7x+14y=35   ...(3)

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

7x+14y=35

7x3y=13
_______________

11y=22

y=2

On 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).

Solve the equations for x and y.

2x3y=7

5x+y=9

A.

3,4

B.

2, -1

C.

2,2

D.

3, -1

#### SOLUTION

Solution : B

Given

2x3y=7...(i)

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

Rearranging (ii), we get  y=95x...(iii)

Substituting (iii) in (i),  we get

2x3(95x)=7

17x=34

x=2.

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

2(2)3y=7

y=1.