Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

Determine whether the system of linear equations $$2x + 3y - 5 = 0, 6x + 9y - 15 = 0$$ has a unique solution, no solutions, or an infinite number of solutions.

A
Infinite number of solutions

B
No solutions

C
Unique solution

D
Cannot be determined

Solution

The equations are
$$\Rightarrow 2x+3y-5=0$$
here, $$ a_{1}=2,b_{1}=3,c_{1}=-5$$

$$\Rightarrow 6x+9y-15=0$$
here, $$ a_{2}=6,b_{2}=9,c_{2}=-15$$

$$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{2}{6}= \frac{1}{3} $$

$$\Rightarrow \displaystyle \frac{b_{1}}{b_{2}}=\frac{3}{9}=\frac{1}{3}$$

$$\Rightarrow \displaystyle \frac{c_{1}}{c_{2}}=\frac{-5}{-15}=\frac{1}{3}$$

$$\Rightarrow \displaystyle \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{c_{1}}{c_{2}}$$

Hence, the given system of equations has an infinite number of solutions.
Question 2
1 / -0

Solve graphically the following pair of linear equations:
$$2x + 3y - 12 = 0, 2x - y - 4 = 0$$.

A
$$x=0, y=1$$

B
$$x=4, y=3$$

C
$$x=3, y=2$$

D
$$x=0, y=5$$

Solution

$$2x+3y-12=0$$
$$\Rightarrow 2x=12-3y$$
$$\Rightarrow x=\cfrac{12-3y}{2}$$
Let, $$y=2$$, then $$x=\cfrac{12-3(2)}{2}=3$$
Similarly, when $$y=0, x=6$$
$$2x-y-4=0$$
$$\Rightarrow 2x=4+y$$
$$\Rightarrow x=\cfrac{4+y}{2}$$
Let, $$y=2$$, then $$x=\cfrac{4+2}{2}=3$$
Similarly, when $$y=0, x=2$$
Plotting the points above, the intersecting point is $$(3,2)$$
$$\therefore x=3,y=2$$
Question 3
1 / -0

Solve the following equations by the substitution method
$$\dfrac {1}{2}(9x+10y)=23, \dfrac {5x}{4}-2y=3$$

A
$$x = 4, y =1$$

B
$$x = 2, y =5$$

C
$$x = 1, y =-1$$

D
$$x = 7, y =-3$$

Solution

The equations are
$$\Rightarrow \frac{1}{2}\left ( 9x+10y \right )=23\Rightarrow 9x+10y=46....eq\left ( 1 \right )$$
$$\Rightarrow x=\frac{46-10y}{9}$$
$$\Rightarrow \frac{5x}{4}-2y=3\Rightarrow 5x-8y=12....eq\left ( 2 \right )$$
Substitute the value of x in eq (2 )
$$\Rightarrow \frac{5\left (46-10y \right )}{9}-8y=12$$
$$\Rightarrow 230-50y-72y=108\Rightarrow -122y=-122\Rightarrow y=1$$
Substitute the value of y in eq (2 )
$$\Rightarrow 5x-8\left (1 \right )=12\Rightarrow 5x=20\Rightarrow x=4$$
$$\Rightarrow x=4,y=1 $$
Question 4
1 / -0

Solve graphically the pair of equations $$x+3y=6$$, and $$3x-5y=18$$. Hence, find the value of $$K$$ if $$7x+3y=K$$

A
$$x=-3, y=1, K=13$$

B
$$x=-8, y=5, K=8$$

C
$$x=1, y=2, K=29$$

D
$$x=6, y=0, K=42$$

Solution

From the graph the point of intersection is $$(x,y)=(6,0)$$.
Substituting $$(x,y)=(6,0)$$ in $$7x+3y=K$$, we get
$$7\times 6+3\times 0=K$$
i.e $$K=42$$
Ans- Option D.
Question 5
1 / -0

If $$2x + y = 23$$ and $$4x - y = 19$$, find the values of $$5y - 2x$$ and $$\dfrac{y}{x} - 2$$.

A
$$36, -\dfrac13$$

B
$$31, -\dfrac57$$

C
$$38, \dfrac67$$

D
None of these

Solution

Given System of Equations are:
$$2x+y = 23 $$ ...$$(1)$$
$$4x-y = 19$$ ...$$(2)$$
Now adding both the equation will have:
$$ 6x = 42$$
$$ x = 7$$
Now substituting $$x = 7$$ in in equation $$(1)$$
$$\Rightarrow 2\left ( 7 \right ) + y = 23$$
$$\Rightarrow 14 + y = 23$$
$$ \Rightarrow y = 9$$
Now substituting $$x = 7$$ and $$y = 9$$ in $$5y - 2x$$
$$\Rightarrow 5\left ( 9 \right ) - 2\left ( 7 \right )$$
$$ \Rightarrow 45-14$$
$$ \Rightarrow 31$$
Now Substituting Now substituting $$x = 7$$ and $$y = 9$$ in $$\frac{y}{x}-2$$
$$ \Rightarrow \dfrac{9}{7}-2$$
$$\Rightarrow \dfrac{9-14}{7}$$
$$\Rightarrow \dfrac{-5}{7}$$
Question 6
1 / -0

Solve the following equations by the substitution method
$$0.04 x + 0.02y = 5, 0.5x - 0.4y = 30$$

A
$$x=27, y=61$$

B
$$x=100, y=50$$

C
$$x=200, y=39$$

D
$$x=54, y=122$$

Solution

The equations are
$$\Rightarrow .04x+.02y=5....eq\left ( 1 \right )$$
$$\Rightarrow x=\frac{5-.02y}{.04}$$
$$\Rightarrow .5x-.4y=30....eq\left (2 \right )$$
Substitute the value of x in eq (2)
$$\Rightarrow \frac{.5\left (5-.02y \right )}{.04}-.4y=30$$
$$\Rightarrow 2.5-.01y-.016y=1.2\Rightarrow -.026y=-1.3\Rightarrow y=50$$
Substitute the value of y in eq (2)
$$\Rightarrow .5x-.4\left (50 \right )=30\Rightarrow .5x=50\Rightarrow x=100$$
$$\Rightarrow x=100,y=50 $$
Question 7
1 / -0

Solve the following equations by the substitution method
$$x = 3y - 19, y = 3x - 23$$

A
$$x = 5, y = 7$$

B
$$x = 11, y = 10$$

C
$$x = 13, y = 7$$

D
$$x = 3, y = 11$$

Solution

The equations are
$$\Rightarrow x=3y-19....eq\left ( 1 \right )$$
$$\Rightarrow y=3x-23....eq\left ( 2 \right )$$
Substitute the   value of x in   eq (2 )
$$\Rightarrow y=3\left (3y-19 \right )-23 $$
$$\Rightarrow y=9y-57-23\Rightarrow -8y=-80\Rightarrow y=10$$
Substitute the value of y in   eq (1 )
$$\Rightarrow x=3\left (10 \right )-19\Rightarrow x=30-19\Rightarrow x=11$$
$$\Rightarrow x=11,y=10  $$
Question 8
1 / -0

Solve the following pair of equations graphically : $$x + y = 4, 3x - 2y =- 3$$
Shade the region bounded by the lines representing the above equations and x-axis

A
$$x = 3, y = 2$$

B
$$x = 1, y = 3$$

C
$$x = 8, y = 2$$

D
$$x = 9, y = 2$$

Solution

Equation (i) is $$x+y=4 $$
i.e. $$y=4-x$$ and
$$3x-2y=-3$$
i.e $$y=\dfrac {(3x+3)}{2}$$
The point of intersection from the graphs is $$(x,y)=(1,3)$$
The required area is a triangle
Ans- Option B.
Question 9
1 / -0

Solve the following equations by the substitution method
$$\dfrac {x+11}{7}+2y=10, 3x=8+\dfrac {y+7}{11}$$

A
$$x = 1, y = -2$$

B
$$x = 8, y = -7$$

C
$$x = 9, y = 2$$

D
$$x = 3, y = 4$$

Solution

The equations are
$$\Rightarrow \dfrac{x+11}{7}+2y=10\Rightarrow x+11+14y=70$$
$$\Rightarrow x+14y=59 ....eq\left ( 1 \right )$$
$$\Rightarrow x=59-14y$$
$$\Rightarrow 3x=8+\dfrac{y+7}{11}\Rightarrow 33x=88+y+7$$
$$\Rightarrow 33x-y=95....eq\left ( 2 \right )$$
Substitute the value of x in eq (2 )
$$\Rightarrow 33\left (59-14y \right )-y=95 $$
$$\Rightarrow 1947-462y-y=95\Rightarrow 463y=1853\Rightarrow y=4 $$
$$\Rightarrow x=59-14y\Rightarrow x=59-14\left (4 \right ) \Rightarrow x=3$$
$$\Rightarrow x=3,y=4$$
Question 10
1 / -0

Solve the following equations by the substitution method
$$11x-8y=27, 3x+5y=-7$$

A
$$x=0, y=1$$

B
$$x=0, y=-5$$

C
$$x=2, y=3$$

D
$$x=1, y=-2$$

Solution

The equations are
$$\Rightarrow 11x-8y=27....eq\left ( 1 \right )$$
$$\Rightarrow x=\frac{27+8}{11}$$
$$\Rightarrow 3x+5y=-7....eq\left ( 2 \right )$$
Substitute the value of x in eq(2 )
$$\Rightarrow \frac{3\left (27+8y \right )}{11}+5y=-7$$
$$\Rightarrow 81+24y+55y=-77\Rightarrow 79y=-158\Rightarrow y=-2$$
Substitute the value of y in eq (2 )
$$\Rightarrow 3x+5\left (-2 \right )=-7\Rightarrow 3x=3\Rightarrow x=1$$
$$\Rightarrow x=1,y=-2 $$

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Pair of Linear Equations in Two Variables Test - 40

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Pair of Linear Equations in Two Variables Test - 40

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Question 1

Infinite number of solutions

No solutions

Unique solution

Cannot be determined

Solution

Question 2

$$x=0, y=1$$

$$x=4, y=3$$

$$x=3, y=2$$

$$x=0, y=5$$

Solution

Question 3

$$x = 4, y =1$$

$$x = 2, y =5$$

$$x = 1, y =-1$$

$$x = 7, y =-3$$

Solution

Question 4

$$x=-3, y=1, K=13$$

$$x=-8, y=5, K=8$$

$$x=1, y=2, K=29$$

$$x=6, y=0, K=42$$

Solution

Question 5

$$36, -\dfrac13$$

$$31, -\dfrac57$$

$$38, \dfrac67$$

None of these

Solution

Question 6

$$x=27, y=61$$

$$x=100, y=50$$

$$x=200, y=39$$

$$x=54, y=122$$

Solution

Question 7

$$x = 5, y = 7$$

$$x = 11, y = 10$$

$$x = 13, y = 7$$

$$x = 3, y = 11$$

Solution

Question 8

$$x = 3, y = 2$$

$$x = 1, y = 3$$

$$x = 8, y = 2$$

$$x = 9, y = 2$$

Solution

Question 9

$$x = 1, y = -2$$

$$x = 8, y = -7$$

$$x = 9, y = 2$$

$$x = 3, y = 4$$

Solution

Question 10

$$x=0, y=1$$

$$x=0, y=-5$$

$$x=2, y=3$$

$$x=1, y=-2$$

Solution

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