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:

$\begin{matrix} 2 x + 3 y - 8 = 0 \\ 3 x - 4 y + 5 = 0 \end{matrix}$

Accepted Answer

We know that if two lines are represented by $\begin{matrix} 2 x + 3 y - 8 = 0 \\ 3 x - 4 y + 5 = 0 \end{matrix}$ 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 $2 x + 3 y - 8 = 0$ and $3 x - 4 y + 5 = 0$ .

So, the value of the coefficients are $a_{1} = 2, b_{1} = 3, c_{1} = - 8, a_{2} = 3, b_{2} = - 4$ and $c_{2} = 5$ .

We get, $\frac{a_{1}}{a_{2}} = \frac{2}{3}$ and $\frac{b_{1}}{b_{2}} = \frac{3}{- 4} = - \frac{3}{4}$ .

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

$\Rightarrow \frac{x}{3 \times 5 - (- 4) \times (- 8)} = \frac{1}{2 \times (- 4) - 3 \times 3}$

$\Rightarrow \frac{x}{15 - 32} = \frac{1}{- 8 - 9}$

$\Rightarrow \frac{x}{- 17} = \frac{1}{- 17}$

$\Rightarrow x = \frac{- 17}{- 17} = 1$

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

$\Rightarrow \frac{y}{- 8 \times 3 - 5 \times 2} = \frac{1}{2 \times (- 4) - 3 \times 3}$

$\Rightarrow \frac{y}{- 24 - 10} = \frac{1}{- 8 - 9}$

$\Rightarrow \frac{y}{- 34} = \frac{1}{- 17}$

$\Rightarrow y = \frac{- 34}{- 17} = 2$

Therefore, the required unique solution is $x = 1$ and $y = 2$ .

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:

$\begin{matrix} 2 x + 3 y - 8 = 0 \\ 3 x - 4 y + 5 = 0 \end{matrix}$

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: 2x+3y-8=03x-4y+5=0

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:

$\begin{matrix} 2 x + 3 y - 8 = 0 \\ 3 x - 4 y + 5 = 0 \end{matrix}$