How do you solve $3x+y=16$ & $2x-3y=-4$ using substitution?

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Answer 1 · 2016-09-05T18:53:04

$x = 4, y = 4$

We have: $3 x + y = 16$ and $2 x - 3 y = - 4$

Let's solve the first equation for $y$ :

$=> y = 16 - 3 x$

Then, let's substitute this value of $y$ into the second equation:

$=> 2 x - 3 (16 - 3 x) = - 4$

$=> 2 x - 48 + 9 x = - 4$

$=> 11 x = 44$

$=> x = 4$

Now, let's substitute this value of $x$ into the equation for $y$ :

$=> y = 16 - 3 (4)$

$=> y = 16 - 12$

$=> y = 4$

Therefore, the solutions to the system of equations is $x = 4$ and $y = 4$ .

How do you solve 3x+y=163x+y=16 & 2x-3y=-42x-3y=-4 using substitution?