Name: Linear Equations With One or More Variables
Uploaded: 2019-07-19
Duration: 11 min 20 s
Description: Let's learn the concept of linear equations, the introduction of linear algebra, the formation of linear equations in one, two and three variables, etc.

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

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

Unique solution

Two intersecting lines

$2 x + 3 y = 5$

$- 4 x - 6 y = 8$

Accepted Answer

The given equations are $2 x + 3 y = 5$ -----(i)

and, $- 4 x - 6 y = 8$ ----- (ii)

Comparing equation (i) with $a_{1} x + b_{1} y = c_{1}$ and equation (ii) with $a_{2} x + b_{2} y = c_{2}$ , we get, $a_{1} = 2, b_{1} = 3, c_{1} = 5$ and $a_{2} = - 4, b_{2} = - 6, c_{2} = 8$ .

Hence, the ratio of coefficients of $x$ is $\frac{a_{1}}{a_{2}} = \frac{2}{- 4} = - \frac{1}{2}$ .

The ratio of coefficients of $y$ is $\frac{b_{1}}{b_{2}} = \frac{3}{- 6} = - \frac{1}{2}$ .

The ratio of the constant terms is $\frac{c_{1}}{c_{2}} = \frac{5}{8}$ .

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 $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ , then, the pair of equations have no solution. The geometric meaning of this is that the system of equations is depicted by two parallel lines

Here, in this case, we observe that $- \frac{1}{2} = - \frac{1}{2}$ . So, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$ . Also, $- \frac{1}{2} \neq \frac{5}{8}$ . So, in this case we have, we have, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Therefore, the lines $2 x + 3 y = 5$ and $- 4 x - 6 y = 8$ have no solution and the system of equations is depicted by two parallel 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

$2 x + 3 y = 5$

$- 4 x - 6 y = 8$
$\frac{a_{1}}{a_{2}} = \frac{2}{- 4} = - \frac{1}{2}$ $\frac{b_{1}}{b_{2}} = \frac{3}{- 6} = - \frac{1}{2}$ $\frac{c_{1}}{c_{2}} = \frac{5}{8}$ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ No solution Two parallel lines

Important Questions on Linear Equations in Two Variables

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