How do I solve this system of linear equations?

Question

So when I graph $2x - y = 6$ and $x + 2y = -2$, I see them intersecting at points $(2,-2)$.

HOWEVER, when I set them equal to each other $(2x - 6 = -(1/2)x -1)$ I don't get $2$ for $x$. Can someone please clarify how to do this?

Answer 1

$\rm Method\, I\,:Substitution$

We have the system:

$$ 2x-y=6\tag 1 \\ x+2y=-2 $$

From $(1)$: $$2x-y=6\rightarrow y =2x-6\tag 2$$

Plugging into the second equation: $$ x+2y=-2 \iff x+2(2x-6)=-2\\ 5x-12=-2 \iff 5x=10 \iff x=2 $$

Plugging this back in $(2)$:

$$y=2x-6\iff y=2(2)-6\iff y=-2$$

$\rm Method\, II\,: Setting\, equal $ (Don't know name for this): $$ 2x-y=6\iff y=2x-6\\ x+2y=-2 \iff y=-\frac x 2 -1\\ 2x-6=-\frac x 2 -1\iff 4x-12=-x-2\\ 5x=10 $$

Answer 2

Your method of solving the system of equations does work: \begin{align*} 2x-6 &=-\frac{1}{2}x-1\\ 2x+\frac{1}{2}x&=6-1\\ \left(2+\frac{1}{2}\right) x &=5\\ \frac{5}{2}x&=5\\ x&=5\cdot \frac{2}{5}\\ x&=2. \end{align*} Then just plug this value of $x$ back into either equation to solve for $y$. If you want to clear the denominator earlier on: \begin{align*} 2x-6 &=-\frac{1}{2}x-1\\ 4x-12&=-x-2\\ 4x+x &=12-2\\ 5x&=10\\ x&=2.\\ \end{align*}

Answer 3

You are correct till : $ 2x - 6 = {{-x}\over{2}} - 1 $

Maybe , you are solving it the wrong way .

$$ 2x - 6 = {{-x}\over{2} } - 1 $$

$$ \implies 2x + {{x}\over{2}} = - 1 + 6 $$

$$ \implies {{5x }\over{2} } = 5 $$

               DIVIDING BY 5 ON BOTH SIDES 

$$ {x\over2} = 1 $$

$$ \implies x = 2 $$