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

Question

5722 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-05T18:53:04

$x = 4, y = 4$ $x = 4, y = 4$

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

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

$\Rightarrow y = 16 - 3 x$ $=> y = 16 - 3 x$

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

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

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

$\Rightarrow 11 x = 44$ $=> 11 x = 44$

$\Rightarrow x = 4$ $=> x = 4$

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

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

$\Rightarrow y = 16 - 12$ $=> y = 16 - 12$

$\Rightarrow y = 4$ $=> y = 4$

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

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