How do you solve by substitution $8 x - y = 4$ $8x - y = 4$ and $5 x + 3 y = 17$ $5x + 3y = 17$ ?

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Answer 1 · 2015-06-20T12:38:44

Answer: $x = 1, y = 4$ $x=1,y=4$

${(8x-y=4),(5x+3y=17) :}$

${(y=8x-4),(5x+3y=17) :}$

Now we can substitute $8x+4$ as $y$ in the second equation:

$5x+3(8x-4)=17$
$5x+24x-12=17$
$29x=29$
$x=1$

Now we substitute $1$ as $y$ in $8x-y=4$

$8*1-y=4$
$y=4$

So we finally get: ${(x=1),(y=4):}$

How do you solve by substitution 8x - y = 48x−y=4 8x - y = 4 and 5x + 3y = 175x+3y=175x + 3y = 17?