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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

(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,

 

 

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

 

 

 

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

 

 

 

 

 

By the method of Substitution:

 

From the equation (i) we get,

 

 

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

 

 

 

 

Substituting this value in equation (v) we get,

 

 

 

 

 

By the method of Elimination.

 

 

 

 

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

 

 

 

When the equation (iii) is substituted in (i) we get,

 

 

 

 

Hence, x = 2 , y = -3

 

By the method of Substitution:

 

From the equation (ii) we get,

 

 

Putting the value obtained from equation (v) in equation (i) we get,

 

 

 

 

When y = -3 is substituted in equation (v) we get,

 

 

Therefore, x = 2 and y = -3