Question

Form the pair of linear equations for the following problem and find their solution by substitution method.

The larger of two supplementary angles exceeds the smaller by $18$ degrees.

• A. $x+y=100$ and $x-y=18$, $x=90$ and $y=16$
• B. $x+y=60$ and $x-y=18$, $x=95$ and $y=70$
• C. $x+y=180$ and $x-y=18$, $x=99$ and $y=81$
• D. None of these

Answer 1

Answer: (C)

$x+y=180$ and $x-y=18$, $x=99$ and $y=81$

Let the supplementary angles be $x$ and $y$, $(x>y)$

$\therefore$ Sum of both angles is ${180}^0$

$\Rightarrow x+y=180$ ...$\left(1\right)$

The larger angle exceeds the smaller by ${18}^0$

$\Rightarrow x-y=18$ ...$\left(2\right)$

$\Rightarrow y=x-18$

Put value of $y$ in (1), we get

$2x=198$

$\Rightarrow x=99$

Now substituting the value of $x$ in equation $\left(1\right)$

$\Rightarrow 99+y=180$

$\Rightarrow y=180-99$

$\Rightarrow y=81$

Hence, $x={99}^0$ and $y={81}^0$

The pair of linear equations for the given problem are

$x+y=180$ and $x-y=18$ .