Question

Help with math?Please? ANYONE

Answer 1

Answer:

(-1, 1)

Step-by-step explanation:

Hi there!

We want to solve the system of equations given as:
2x-3y=-5

3x+y=-2

Let's solve this equation by substitution, where we will set one variable equal to an expression containing the other variable, substitute the expression as the variable that it equals, solve for the other variable (the variable that the expression contains), and then use the value of the solved variable to find the value of the first variable
In the second equation, we have y by itself; therefore, if we subtract 3x from both sides, then we will get an expression that y is equal to.

So subtract 3x from both sides

y=-3x-2

Now substitute -3x-2 as y in the first equation.
It will look something like this:

2x - 3(-3x-2)=-5

Now do the distributive property.

2x+9x+6=-5

Combine like terms

11x+6=-5

Subtract 6 from both sides

11x=-11

Divide both sides by 11

x=-1

Now substitute -1 as x in the equation y=-3x-2 to solve for y:

y=-3(-1)-2

multiply

y=3-2

Subtract

y=1

The answer is x=-1, y=1; this can also be written as an ordered pair, which would be (-1, 1)
Hope this helps!
If you would like to see another problem for additional practice, take a look here: brainly.com/question/19212538

Answer 2

$\begin{cases} 2x-3y=-5\\ 3x+y=-2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 2x-3y=-5\implies 2x=3y-5\implies x=\cfrac{3y-5}{2} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 2nd equation}}{3\left( \cfrac{3y-5}{2} \right)+y=-2}\implies \cfrac{3(3y-5)}{2}+y=-2 \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( \cfrac{3(3y-5)}{2}+y \right)}=2(-2)\implies 3(3y-5)+2y=-4 \\\\\\ 9y-15+2y=-4\implies 11y-15=-4\implies 11y=11$

$y=\cfrac{11}{11}\implies \blacktriangleright y=1 \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{3y-5}{2}}\implies x=\cfrac{3(1)-5}{2}\implies x=\cfrac{-2}{2}\implies \blacktriangleright x=-1 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (-1~~,~~1)~\hfill$