Question

Can someone help me out with this? I’m not sure what I’m doing wrong.

Answer 1

Answer:

Step-by-step explanation:

This is a system of inequalities such that:

$\left \{ {{2x+3y\geq 2} \atop {3x-4y\leq 3}} \right.$

Let's start by solving for y for both equations:

Equation 1:

$2x+3y\geq 2\\\\3y\geq -2x+2\\\\y \geq \frac{-2x+2}{3}$

Equation 2:

$3x-4y\leq 3\\\\-4y\leq -3x+3\\\\y\geq -\frac{(-3x+3)}{4}$

Now if we substitute the 2nd y into the first equation we obtain:

$2x+3(\frac{3x-3}{4}) \geq 2\\\\2x+\frac{9x-9}{4}\geq 2\\\\\frac{8x+9x-9}{4}\geq 2\\\\17x-9\geq 8\\\\17x\geq 17\\\\x\geq 1$

Now we will solve for the second equation using the first result of y and we obtain:

$3x-4(\frac{-2x+2}{3}\leq 3\\\\3x+\frac{8x-8}{3}\leq 3\\\\\frac{9x+8x-8}{3}\leq 3\\\\17x-8\leq 9\\\\17x\leq 17\\\\x\leq 1$

And so our solution for the system of equations is:

$x \leq 1\\and \\y\geq \frac{-2x+2}{3}$

As well as:

$x>1\\and\\y\geq \frac{3x-3}{4}$