Solve the following pair of linear equations by the elimination method and the substitution method:(i) 3x – 5y – 4 = 0 and 9x = 2y + 7 (ii) x/2+ 2y/3 = -1 and x-y/3 = 3

CBSE | Class 10 | Exercise 3.4 | Pair of Linear Equations In Two Variables

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By the method of elimination:

$<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno"> </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b0ede45f96e02ef7d5e2b4db88f483bd_l3.png" height="201" width="831" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \begin{array}{*{35}{l}} 3x\text{ }\text{ }5y\text{ }\text{ }4\text{ }=\text{ }0\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text{ }\left( i \right) \\ ~ \\ 9x\text{ }=\text{ }2y\text{ }+\text{ }7 \\ ~ \\ \end{array} \\ & 9x\text{ }\text{ }2y\text{ }\text{ }7\text{ }=\text{ }0\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left( ii \right) \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>$

When the equation (i) is multiplied by 3 we get,

$9x\text{ }\text{ }15y\text{ }\text{ }12\text{ }=\text{ }0\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left( iii \right)$

When the equation (iii) is subtracted from equation (ii) we get,

$13y\text{ }=\text{ }-5$

$y\text{ }=\text{ }-5/13\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\left( iv \right)$

When equation (iv) is substituted in equation (i) we get,

$3x\text{ }+25/13\text{ }-4=0$

$3x\text{ }=\text{ }27/13$

$x\text{ }=9/13$

$\therefore x\text{ }=\text{ }9/13\text{ }and\text{ }y\text{ }=\text{ }-5/13$

By the method of Substitution:

From the equation (i) we get,

$x\text{ }=\text{ }\left( 5y+4 \right)/3\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text{ }\left( v \right)$

Putting the value (v) in equation (ii) we get,

$9\left( 5y+4 \right)/3\text{ }-2y\text{ }-7=0$

$13y\text{ }=\text{ }-5$

$y\text{ }=\text{ }-5/13$

Substituting this value in equation (v) we get,

$x\text{ }=\text{ }\left( 5\left( -5/13 \right)+4 \right)/3$

$x\text{ }=\text{ }9/13$

$\therefore x\text{ }=\text{ }9/13,\text{ }y\text{ }=\text{ }-5/13$

$\left( iv \right)\text{ }x/2\text{ }+\text{ }2y/3\text{ }=\text{ }-1\text{ }and\text{ }x-y/3\text{ }=\text{ }3$

By the method of Elimination.

$3x\text{ }+\text{ }4y\text{ }=\text{ }-6\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\text{ }\left( i \right)$

$x-y/3\text{ }=\text{ }3$

$3x\text{ }\text{ }y\text{ }=\text{ }9\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\text{ }\left( ii \right)$