Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 50

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Weekly Quiz Competition

Question 1
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Father says to son "I am $$5$$ times as old as you were when i was as old as you are". If sum of their ages is $$48$$, what is the age of the father?

A
$$20$$

B
$$9$$

C
$$45$$

D
$$40$$

Solution

Let the present age of father and son is x and y
Then as per question $$x+y=48$$ ........................(1)
Given $$x=5y$$ .......................................................(2)
substituting equation 2 in equation 1 we get
$$5y+y=48$$
$$6y=48$$
$$y=8$$.........................................................................(3)
Putting value of y in eqn 2 we get $$x=5(8)=40$$
We get $$x=40$$
Then age of father is $$40$$ years
Question 2
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Graph the equation by determining the missing values needed to plot the ordered pairs.
$$y+x=4;$$
$$\left( 1, y_1\right) ,\left( 4, y_2 \right) ,\left( 3, y_3\right) $$

A

B

C

D

Solution

We have given line, $$y+x=4\Rightarrow y=4-x$$
Now, at $$x=1, y=4-1=3$$
At $$x=4, y=4-4=0$$ and at $$x=3, y=4-3=1$$
Hence, given points on the line are $$(1,3),(4,0),(3,1)$$.
Clearly these points lying on the graph of line in option B.
Hence option B is correct choice.
Question 3
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Find the value of $$x$$ and $$y$$ using substitution method:
$$x- y = 2$$ and $$2x - y = 9$$

A
$$7$$ and $$7$$

B
$$7$$ and $$5$$

C
$$-7$$ and $$-5$$

D
$$5$$ and $$7$$

Solution

$$x- y = 2$$ ------- $$(1)$$
$$2x - y = 9$$ ------ $$(2)$$
From equation $$(1)$$ :
$$y = x - 2$$
Substitute the value of $$y$$ in equation $$(2)$$:
$$2x - y = 9$$
$$2x -x + 2 = 9$$
$$x = 9 -2$$
$$x = 7$$
Now, Substitute $$x = 7$$ in equation $$(1)$$:
$$x- y = 2$$
$$7- y = 2$$
$$y = 7 -2$$
$$y = 5$$
Therefore the solution is: $$x = 7$$ and $$y = 5$$
Question 4
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Using substitution method find the value of x and y:
$$2x - 3y = 2$$ and $$5x - 6y = 2$$

A
-4 and -1

B
-2 and 10

C
-2 and -11

D
-2 and -2

Solution

$$2x- 3y = 2$$ ------- (1)
$$5x- 6y = 2$$------ (2)
From equation 1:
$$2x- 2 = 3y$$
$$y =\dfrac{2x-2}{3}$$
Substitute the value of y in equation 2:
$$5x- 6y = 2$$
$$5x- 6(\dfrac{2x-2}{3}) = 2$$
$$5x- 2(2x- 2) = 2$$
$$5x- 4x + 4 = 2$$
$$x = 2- 4$$
$$x = -2$$
Now, Substitute $$x = -2$$ in equation 1:
$$5x- 6y = 2$$
$$5(-2)- 6y = 2$$
$$-10 -2 = 6y$$
$$6y = -12$$
$$y = \dfrac{-12}{6} = -2$$
Therefore the solution is: $$x = -2$$ and $$y = -2$$
Question 5
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Use the method of substitution to solve the equations
$$x+2y=-4$$ and $$4x+5y=2$$

A
$$\dfrac{-16}{3}$$ and $$\dfrac{-2}{3}$$

B
$$\dfrac{-16}{3}$$ and $$\dfrac{2}{3}$$

C
$$8$$ and $$-6$$

D
$$\dfrac{16}{3}$$ and $$\dfrac{2}{3}$$

Solution

$$x+2y=-4$$
$$\implies x=-4-2y$$
Substituting $$ x=-4-2y$$ in
$$4x+5y=2$$
$$\implies 4(-4-2y)+5y=2$$
$$\implies -16-8y+5y=2$$
$$\implies -16-3y=2$$
$$\implies -16-2=3y$$
$$\implies y=\dfrac{-18}{3}=-6$$
Substituting $$y=-6$$ in $$ x=-4-2y$$, we get,
$$x=-4-2(-6)$$
$$=-4+12=8$$
$$\therefore x=8, y=-6$$
Question 6
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Solve the equation using substitution method:
$$3x-5y =11$$ and $$+10y+6x=7$$

A
$$\left(\frac{29}{12}, \frac{3}{4}\right)$$

B
$$\left(\frac{3}{4}, \frac{29}{12}\right)$$

C
$$\left(\frac{-29}{12}, \frac{3}{4}\right)$$

D
$$\left(\frac{29}{12}, \frac{-3}{4}\right)$$

Solution

$$3x-5y =11$$ ---------- (1)
$$+10y+6x=7$$ ---------(2)
From equation (1)
$$3x-5y=11$$
$$\frac{3x-11}{5}=y$$
Substitute the value of $$y$$ in equation (2)
$$+10\left(\frac{3x-11}{5}\right)+6x = 7$$
$$+6x-22+6x = 7$$
$$12x=7+22$$
$$x=\frac{29}{12}$$
Substitute the value of $$x$$ in equation (1)
$$3x-5y = 11$$
$$3\left(\frac{29}{12}\right) - 5y = 11$$
$$29-20y = 44$$
$$29-44=20y$$
$$\frac{-153}{20} = y$$
$$y=\frac{-3}{4}$$
therefore solution is $$x=\frac{29}{12}$$ and $$y=\frac{-3}{4}$$
Question 7
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Using substitution method find the value of $$x$$ and $$y:$$
$$x + 4y = -4$$ and $$2x- 3y = 2$$

A
$$\dfrac{-4}{11}$$ and $$\dfrac{-10}{11}$$

B
$$\dfrac{-2}{11}$$ and $$\dfrac{10}{11}$$

C
$$\dfrac{-2}{11}$$ and $$\dfrac{-10}{11}$$

D
$$\dfrac{2}{11}$$ and $$\dfrac{-2}{11}$$

Solution

$$x + 4y = -4$$ ------- $$(1)$$
$$2x- 3y = 2$$ ------ $$(2)$$
From equation $$1:$$
$$4y = -4 -x$$
$$y =\dfrac{-4-x}{4}$$
Substitute the value of $$y$$ in equation $$2:$$
$$2x- 3y = 2$$
$$2x- 3\left(\dfrac{-4-x}{4}\right) = 2$$
$$8x + 12 + 3x = 8$$
$$11x = 8-12$$
$$11x = -4$$
$$x = \dfrac{-4}{11}$$
Now, Substitute $$x =\dfrac{-4}{11}$$ in equation $$1:$$
$$x + 4y = -4$$
$$\dfrac{-4}{11} + 4y = -4$$
$$-4 + 44y = -44$$
$$44y = -44 + 4$$
$$44y = -40$$
$$y = \dfrac{-40}{44} \\= \dfrac{-10}{11}$$
Therefore the solution is: $$x = \dfrac{-4}{11}$$ and $$y = \dfrac{-10}{11}$$
Question 8
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Using substitution method find the value of x and y:
$$3x+5y$$ $$= 1$$ and $$4x + 5y $$$$= 2$$

A
$$-1$$ and $$\dfrac{-2}{5}$$

B
$$-1$$ and $$\dfrac{2}{5}$$

C
$$1$$ and $$\dfrac{-2}{5}$$

D
$$1$$ and $$\dfrac{2}{5}$$

Solution

$$3x+5y=1$$
$$\implies 3x=1-5y$$
$$\implies x=\dfrac{1-5y}{3}$$
Now, Substituting $$x=\dfrac{1-5y}{3}$$ in
$$4x+5y=2$$

$$\implies 4(\dfrac{1-5y}{3})+5y=2$$

$$\implies \dfrac{4-20y+15y}{3}=2$$

$$\implies 4-5y=6$$
$$\implies -2=5y$$
$$\implies y=\dfrac{-2}{5}$$

Substituting $$y=\dfrac{-2}{5}$$ in $$x=\dfrac{1-5y}{3}$$
$$x=\dfrac{1-5(\dfrac{-2}{5})}{3}$$
$$x=\dfrac{1+2}{3}=1$$
$$\therefore x=1, y=\dfrac{-2}{5}$$
Question 9
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Using substitution method find the value of x and y:
$$4x + 9y = 5$$ and $$-5x + 3y = 8$$

A
$$-1$$ and $$\dfrac{-2}{5}$$

B
$$-1$$ and $$\dfrac{2}{5}$$

C
$$-1$$ and $$1$$

D
$$1$$ and$$\dfrac{2}{5}$$

Solution

$$4x+9y=5$$
$$\implies 4x=5-9y$$
$$\implies x=\dfrac{5-9y}{4}$$
Substituting $$x=\dfrac{5-9y}{4}$$ in
$$-5x+3y=8$$, we get,
$$-5(\dfrac{5-9y}{4})+3y=8$$
$$\implies \dfrac{-25+45y+12y}{4}=8$$
$$\implies -25+45y+12y=8\times 4$$
$$\implies -25+57y=32$$
$$\implies 57y=57$$
$$\implies y=1$$
Substituting $$y=1$$in $$x=\dfrac{5-9y}{4}$$
$$x=\dfrac{5-9(1)}{4}$$
$$\implies x=\dfrac{-4}{4}$$
$$\implies x=-1$$
$$\therefore x=-1, y=1$$
Question 10
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Using substitution method find the value of x and y:
$$x - 9y = 18$$ and $$-x + 3y = -15$$

A
$$-13.5 $$ and $$-3.5$$

B
$$-13.5$$ and $$2.5$$

C
$$13.5$$ and $$-2.5$$

D
$$13.5$$ and $$-0.5$$

Solution

$$x-9y=18$$
$$\implies x=18+9y$$
Substituting $$ x=18+9y$$ in
$$-x+3y=-15$$, we get,
$$ -(18+9y)+3y=-15$$
$$\implies -18-9y+3y=-15$$
$$\implies -6y=-15+18$$
$$\implies -6y=3$$
$$\implies y=-0.5$$
Substituting $$y=-0.5$$ in $$ x=18+9y$$
$$x=18+9(-0.5)$$
$$\implies x=18-4.5$$
$$\implies x=13.5$$
$$\therefore x=13.5, y=-0.5$$