# How do you solve the system y= -2x + 2 and 6x + 2y = 3?

Nov 22, 2015

$x = - \frac{1}{2}$, $y = 3$

#### Explanation:

Substitute in the second equation the expression for $y$ given by the first, i.e. $y = - 2 x + 2$, obtaining

$6 x + 2 \left(- 2 x + 2\right) = 3$

Expanding, you have

$6 x - 4 x + 4 = 3 \setminus \iff 2 x = - 1 \setminus \iff x = - \frac{1}{2}$.

Once $x$ is known, you can obtain $y$ from the foretold relation

$y = - 2 x + 2 \setminus \iff y = - 2 \left(- \frac{1}{2}\right) + 2 = 1 + 2 = 3$

Nov 22, 2015

$\textcolor{b l u e}{\text{Solution: } y = 1 , \textcolor{w h i t e}{\times x} x = - \frac{1}{2}}$

#### Explanation:

Given: $y = - 2 x + 2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$
$\textcolor{w h i t e}{\times \times} 6 x + 2 y = 3. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$

I am going change equation (2) into the format of $y =$something

Then I can equate both equations to each other through $y$
removing that variable. We will then only have $x$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Consider equation (2) "color(green)("Method shown in detail}}$

Subtract $\textcolor{b l u e}{6 x}$ from both sides isolating the y-term

$\textcolor{b r o w n}{\left(6 x + 2 y\right) \textcolor{b l u e}{- 6 x} = \left(3\right) \textcolor{b l u e}{- 6 x}}$

$\textcolor{g r e e n}{\text{The brackets are there only to show what is being altered or}}$
$\textcolor{g r e e n}{\text{grouped to make things easier to see what is happening}}$

$\left(6 x - 6 x\right) + 2 y = 3 - 6 x$

But $6 x - 6 x = 0$ giving

$0 + 2 y = 3 - 6 x$

$2 y = 3 - 6 x$
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$\textcolor{b l u e}{\text{Divide both sides by 2 so that we have y on its own}}$
Not that $\div i \mathrm{de} 2 \to \times \frac{1}{2}$

color(brown)((2y) color(blue)( times 1/2) = (3-6x)color(blue)(times 1/2)

$\frac{2}{2} \times y = \frac{3}{2} - \frac{6}{2} x$

$y = - 3 x + \frac{3}{2.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({2}_{a}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Combining equations the remove the variable } y}$

$E q u a t i o n \left(1\right) = y = E q u a t i o n \left({2}_{a}\right)$

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

$- 2 x + 2 = - 3 x + \frac{3}{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Collecting like terms")color(green)(" Less detail now}}$

$3 x - 2 x = \frac{3}{2} - 2$

$x = - \frac{1}{2.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Substitute (3) into (1) to find y}}$

$y = - 2 x + 2 \to y = \left(- 2\right) \left(\frac{1}{2}\right) + 2$

$y = 1$
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$\textcolor{b l u e}{\text{Solution: } y = 1 , x = - \frac{1}{2}}$