Free Pair of Linear Equations in Two Variables 02 Practice Test - 10th Grade 

We have,

$2x+y=7$    ...(1)
$3x+2y=12$...(2)

Multiply equation (1) by 2, we get: 

$2(2x+y)=2\left(7\right)$
$\Rightarrow 4x+2y=14$...(3)

Subtracting (2) from (3) we get ,

$x=2$
Substituting the value of $x$ in (1) we get,
$2\left(2\right)+y=7⟹y=3$

Thus, the solution for the given pair of linear equations is (2,3).

Let's assume that the speed of the boat in still water is x km/hr and speed of the stream is y km/hr.

So, the speed of the boat in upstream will be (x-y) km/hr.

Similarly, the speed of the boat downstream will be (x+y) km/hr.
We know $\text{time}=\left(\frac{\text{distance}}{\text{speed}}\right)$.

Using the above formula we can form the equations in two variables.

Taking the first case,

$\frac{30}{\text{x - y}}+\frac{44}{\text{x + y}}=14$.

Taking the second case,

$\frac{60}{\text{x - y}}+\frac{55}{\text{x + y}}=25$.

Now, we have the equations in two variables but the equations are not linear.

So, we will assume  $\frac{1}{\text{x - y}}=\text{u}$  and  $\frac{1}{\text{x + y}}=\text{v}$.

So on substituting $\text{u}$ and $\text{v}$ in the above two equations, we get

$30\text{u}+44\text{v}=14$   ...(1)

$60\text{u}+55\text{v}=25$   ...(2)

We can solve the above two equations using the elimination method.

$60\text{u}+88\text{v}=28$   ...(3)
(by multiplying equation (1) by 2)

On subtracting equation (2) from (3), we get $\text{v}=\frac{1}{11}$

On substituting $\text{v}$ in equation (2) we get $\text{u}=\frac{1}{3}$

Now as we have assumed 

$\frac{1}{\text{x - y}}=\text{u}$              and                $\frac{1}{\text{x + y}}=\text{v}$ 

On substituting the values of $\text{u}$ and $\text{v}$,

we get a pair of linear equations in $\text{x and y}$

$\text{x - y}=3$...(4)

$\text{x + y}=11$...(5)

On adding (5) from (4), we have 

$2x=14$

$x=7$ 

On subsituting the value of x in $x-y=3$,  we get y = 4.

So, the speed of the boat in still water is 7 km/hr and the speed of the stream is 4 km/hr.

Given equations gives infinitely many solutions if,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

The given linear equations are:
$3x+ky=9;6x+4y=18.$
 $\Rightarrow a_1=3,b_1=k,c_1=-9$  and   $a_2=6,b_2=4,c_2=-18$

$\Rightarrow \frac{3}{6}=\frac{k}{4}=\frac{-9}{-18}$

$\Rightarrow \frac{1}{2}=\frac{k}{4}$ 

$\Rightarrow k=2$

Take first two components,

$\frac{x+y-8}{2}=\frac{x+2y-14}{8}$

$\Rightarrow 8(x+y-8)=2(x+2y-14)$
$\Rightarrow 8x+8y-64=2x+4y-28$
$\Rightarrow 6x+4y-36=0$

$\Rightarrow 3x+2y-18=0.....\left(i\right)$

Take last two components, 

$\frac{x+2y-14}{8}=\frac{3x+y-12}{11}$

$\Rightarrow 11(x+2y-14)=8(3x+y-12)$

$\Rightarrow 11x+22y-154=24x+8y-96$

$\Rightarrow -13x+14y-58=0.....\left(ii\right)$

On multiplying equation $\left(i\right)$ by 7, we get

$\Rightarrow 21x+14y-126=\mathrm{0...}\left(iii\right)$,

On subtracting equation (ii) from (iii), we get

$\Rightarrow 34x=68$

$\Rightarrow x=2$

On substituting value of $x=2$ in equation $\left(i\right)$, we get

$3\times 2+2y-18=0$

$\Rightarrow y=6$

$\therefore$ The solution is (2, 6).

In the given graph, lines of the equations 2x + 4y - 12 = 0 and x + 2y - 4 = 0 are parallel to each other which gives no solution; hence it is an inconsistent pair of linear equation.

Let, the cost of a bike be $x$.
The cost of scooter be $y$.
According to question $x=2y$

Therefore linear equation will be:
$\Rightarrow x-2y=0$

If the point (1,3k) lies on the given equation, then it should satisfy the equation.

Substituting the values of x and y in the given equation:

$\Rightarrow k\left(1\right)+4\left(3k\right)=26$

$\Rightarrow 13k=26$

$\Rightarrow k=2$

Thus, the value of $k$ is 2.

Option A:

$x+y=–1-----\left(i\right)$

$2x+3y=–5----\left(ii\right)$ 

$\left(i\right)\times 2\to 2x+2y=-2----\left(iii\right)$ 

Solving by elimination$\left(\right(iii)-(ii\left)\right)$ we get,

$x=2andy=-3.$

Option D:

$x-4y=14----\left(i\right)$

$5x-y=13-----\left(ii\right)$
$\left(ii\right)\times 4\to 20x-4y=52----\left(iii\right)$

Solving by elimination method $\left(\right(iii)-(i\left)\right)$ we get,

$x=2andy=-3$

Let starting salary be ₹ x
and annual increment be ₹ y
According to the first condition:
x + 4y = 15000 ......(i)
According to the second condition:
x + 10y = 18000 ......(ii)
Subtracting (i) from (ii), we get
6y = 3000
y = 500
Substituiting y = 500 in equation (i), we get
x + (4$\times$500) = 15000
x = 15000 - 2000
x = 13000
Hence starting salary is ₹ 13000 and annual increment is ₹ 500.

Let the fraction be $\frac{x}{y}$

$\therefore \frac{x+2}{y-1}=\frac{2}{3}and\frac{x+1}{y+2}=\frac{1}{3}$

$\Rightarrow 3x-2y=-8....\left(i\right)$
$and3x-y=-\mathrm{1......}\left(ii\right)$

$\Rightarrow 3x-2y=-8$
$-(3x-y=-1)$

$\Rightarrow 3x-2y=-8$
$-3x+y=1$
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$-y=-7$
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$\Rightarrow y=7$

Substituting y = 7 in equation (ii), we get
$3x-\left(14\right)=-8$
$3x=-8+14=6$
$\therefore x=2$

$\Rightarrow x=2andy=7$