Question

Complete the following table:

System of equations

Ratio of coefficients of

x

Ratio of coefficients of

y

Ratio of constant terms

Relation between ratios

Solutions of equations

Geometrical interpretations

$2 x - 3 y = 1$

$2 x - 4 y = - 4$

\frac{a_{1}}{a_{2}} = \frac{2}{2} = 1

\frac{b_{1}}{b_{2}} = \frac{- 3}{- 4} = \frac{3}{4}

\frac{c_{1}}{c_{2}} = \frac{1}{- 4} = - \frac{1}{4}

\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}

Unique solution

Two intersecting lines

$x + 2 y = 5$

$2 x - 3 y = - 4$

Accepted Answer

The given equations are $x + 2 y = 5$ -----(i)

and, $2 x - 3 y = - 4$ ----- (ii)

Comparing equation (i) with $2 x - 3 y = 1$ and equation (ii) with $a_{2} x + b_{2} y = c_{2}$ , we get, $a_{1} = 1, b_{1} = 2, c_{1} = 5$ and $a_{2} = 2, b_{2} = - 3, c_{2} = - 4$ .

Hence, the ratio of coefficients of $x$ is $\frac{a_{1}}{a_{2}} = \frac{1}{2}$ .

The ratio of coefficients of $y$ is $\frac{b_{1}}{b_{2}} = \frac{2}{- 3} = - \frac{2}{3}$ .

The ratio of the constant terms is $\frac{c_{1}}{c_{2}} = \frac{5}{- 4} = - \frac{5}{4}$ .

We know that if the ratio of the coefficients of the equations $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ are related as $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , then, the pair of equations have an unique solution and the two lines are intersecting lines.

Here, in this case, we observe that $\frac{1}{2} \neq - \frac{2}{3}$ . So, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ .

Therefore, the lines $x + 2 y = 5$ and $2 x - 3 y = - 4$ have an unique solution and they are intersecting lines.

Hence, the completed table is given below:

System of equations Ratio of coefficients of $x$ Ratio of coefficients of $y$ Ratio of constant terms Relation between ratios Solutions of equations Geometrical interpretations

$2 x - 3 y = 1$

$2 x - 4 y = - 4$
$\frac{a_{1}}{a_{2}} = \frac{2}{2} = 1$ $\frac{b_{1}}{b_{2}} = \frac{- 3}{- 4} = \frac{3}{4}$ $\frac{c_{1}}{c_{2}} = \frac{1}{- 4} = - \frac{1}{4}$ $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ Unique solution Two intersecting lines

$x + 2 y = 5$

$2 x - 3 y = - 4$
$\frac{a_{1}}{a_{2}} = \frac{1}{2}$ $\frac{b_{1}}{b_{2}} = \frac{2}{- 3} = - \frac{2}{3}$ $\frac{c_{1}}{c_{2}} = \frac{5}{- 4} = - \frac{5}{4}$ $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ Unique solution Two intersecting lines

Chhattisgarh Board Solutions for Exercise 8: Do this

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