Question

Solve the following equations by the substitution method

$0.04x+0.02y=5,0.5x-0.4y=30$

• A. $x=27,y=61$
• B. $x=100,y=50$
• C. $x=200,y=39$
• D. $x=54,y=122$

Answer 1

The correct option is B $x=100,y=50$

The equations are
$\Rightarrow .04x+.02y=\mathrm{5....}eq\left(1\right)$
$\Rightarrow x=\frac{5-.02y}{.04}$
$\Rightarrow .5x-.4y=\mathrm{30....}eq\left(2\right)$
Substitute the value of x in eq (2)
$\Rightarrow \frac{.5(5-.02y)}{.04}-.4y=30$
$\Rightarrow 2.5-.01y-.016y=1.2\Rightarrow -.026y=-1.3\Rightarrow y=50$
Substitute the value of y in eq (2)
$\Rightarrow .5x-.4\left(50\right)=30\Rightarrow .5x=50\Rightarrow x=100$
$\Rightarrow x=100,y=50$