Name: Fractions Using Cross Multiplication
Uploaded: 2019-07-19
Duration: 3 min 47 s
Description: Learn how to compare two fractions with each other by using the fairly straightforward method of cross-multiplication. Master this concept now by watching th...

Question

Examine whether the following pair of linear equations has a unique solution, no solution, or infinitely many solutions or not? In case there is a unique solution, find it by using the cross-multiplication method:

$\begin{array}{l} x - 3 y = 7 \\ x - y = 5 \end{array}$

Accepted Answer

We know that if two lines are represented by $\begin{array}{l} x - 3 y = 7 \\ x - y = 5 \end{array}$ and $a_{2} x + b_{2} y + c_{2} = 0$ , then the lines have

i) a unique solution, if $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

ii) no solution, if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ and

iii) infinitely many solutions, if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ .

Here, the given lines are $x - 3 y = 7 \Rightarrow x - 3 x - 7 = 0$ and $x - y = 5 \Rightarrow x - y - 5 = 0$ .

So, the value of the coefficients are $a_{1} = 1, b_{1} = - 3, c_{1} = - 7, a_{2} = 1, b_{2} = - 1$ and $c_{2} = - 5$ .

We get, $\frac{a_{1}}{a_{2}} = \frac{1}{1} = 1$ and $\frac{b_{1}}{b_{2}} = \frac{- 3}{- 1} = 3$ .

So, in this case, we observe that $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ .

Hence, the given pair of linear equations has a unique solution.

Now, by using the cross-multiplication method, the unique solution is given by $\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$ .

So, $\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

$\Rightarrow \frac{x}{(- 3) \times (- 5) - (- 1) \times (- 7)} = \frac{1}{1 \times (- 1) - 1 \times (- 3)}$

$\Rightarrow \frac{x}{15 - 7} = \frac{1}{- 1 + 3}$

$\Rightarrow \frac{x}{8} = \frac{1}{2}$

$\Rightarrow x = \frac{8}{2} = 4$ .

And, $\frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

$\Rightarrow \frac{y}{- 7 \times 1 - (- 5) \times 1} = \frac{1}{1 \times (- 1) - 1 \times (- 3)}$

$\Rightarrow \frac{y}{- 7 + 5} = \frac{1}{- 1 + 3}$

$\Rightarrow \frac{y}{- 2} = \frac{1}{2}$

$\Rightarrow y = \frac{- 2}{2} = - 1$

Therefore, the required unique solution is $x = 4$ and $y = - 1$ .

Examine whether the following pair of linear equations has a unique solution, no solution, or infinitely many solutions or not? In case there is a unique solution, find it by using the cross-multiplication method:

$\begin{array}{l} x - 3 y = 7 \\ x - y = 5 \end{array}$

Important Questions on Pair of Linear Equations in Two Variables

Learn and Prepare for any exam you want !

Examine whether the following pair of linear equations has a unique solution, no solution, or infinitely many solutions or not? In case there is a unique solution, find it by using the cross-multiplication method: x-3y=7x-y=5

Important Questions on Pair of Linear Equations in Two Variables

Learn and Prepare for any exam you want !

Examine whether the following pair of linear equations has a unique solution, no solution, or infinitely many solutions or not? In case there is a unique solution, find it by using the cross-multiplication method:

$\begin{array}{l} x - 3 y = 7 \\ x - y = 5 \end{array}$