How do you solve $0.5 x + 0.25 y = 36$ $0.5x+0.25y=36$ and $y + 18 = 16 x$ $y+18=16x$ using substitution?

Question

2978 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-05T06:17:33

$x = 5$ $x = 5$
$y = 62$ $y = 62$

In either equation, express one of the variables in terms of the other terms.

$[1] 0.5 x + 0.25 y = 36$ $[1] 0.5x + 0.25y = 36$
$[2] y + 18 = 16 x$ $[2] y + 18 = 16x$

If we solve for $y$ $y$ in $[2]$ $[2]$ , we have

$[3] \Rightarrow y = 16 x - 18$ $[3] => y = 16x - 18$

Now, substitute this relation into $[1]$ $[1]$

$[1] 0.5 x + 0.25 y = 36$ $[1] 0.5x + 0.25y = 36$

$\Rightarrow 0.5 x + 0.25 (16 x - 18) = 18$ $=> 0.5x + 0.25(16x - 18) = 18$

Multiplying the entire equation by 4, we have

$\Rightarrow 4 (0.5 x + 0.25 (16 x - 18) = 18)$ $=> 4(0.5x + 0.25(16x - 18) = 18)$

$\Rightarrow 2 x + 1 (16 x - 18) = 72$ $=> 2x + 1(16x - 18) = 72$
$\Rightarrow 2 x + 16 x - 18 = 72$ $=> 2x + 16x - 18 = 72$

$\Rightarrow 18 x = 90$ $=> 18x = 90$
$\Rightarrow x = 5$ $=> x = 5$

Using $[3]$ $[3]$ (or $[1]$ $[1]$ or $[2]$ $[2]$ ), solve for $y$ $y$

$y = 16 x - 18$ $y = 16x - 18$
$\Rightarrow y = 16 (5) - 18$ $=> y = 16(5) - 18$

$\Rightarrow y = 80 - 18$ $=> y = 80 - 18$

$\Rightarrow y = 62$ $=> y = 62$

How do you solve 0.5x+0.25y=360.5x+0.25y=36 and y+18=16xy+18=16x using substitution?