How do you solve the system of equations $x + 2 y = 3$ $x+ 2y = 3$ and $- 2 x + y = 1$ $- 2x + y = 1$ ?

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How do you solve the system of equations $x + 2 y = 3$ $x+ 2y = 3$ and $- 2 x + y = 1$ $- 2x + y = 1$ ?

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Answer 1 · 2017-07-30T22:04:38

Pick either one of the equations and make either $x$ $x$ or $y$ $y$ the subjects of the formula and substitute the value of either $y$ $y$ or $x$ $x$ into the other equation.

Explanation:

I.e. $x = 3 - 2 y$ $" "x=3-2y" "$ (from first equation)

Now substitute $x$ $x$ above into second equation:

$- 2 (3 - 2 y) + y = 1$ $-2(3-2y)+y=1$

$- 6 + 4 y + y = 1$ $-6+4y+y=1$

$- 6 + 5 y = 1$ $-6+5y=1$

$5 y = 7$ $5y=7$

Therefore

$y = \frac{7}{5}$ $y=7/5$

Now substitute the $y$ $y$ -value into the equation to find $x$ $x$ :

$- 2 x + \frac{7}{5} = 1$ $-2x+7/5=1$

$- 2 x = \frac{5}{5} (= 1) - \frac{7}{5}$ $-2x=5/5(=1) -7/5$

$x = \frac{1}{5}$ $x=1/5$

Thus, $x$ $x$ and $y$ $y$ for the 2nd equation $= \frac{1}{5}$ $=1/5$ and $= \frac{7}{5}$ $=7/5$ .

You must again substitute either into the 1st equation to find those values. Just remember to check the answers!

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How do you solve the system of equations $x + 2 y = 3$ $x+ 2y = 3$ and $- 2 x + y = 1$ $- 2x + y = 1$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve the system of equations x+ 2y = 3x+2y=3x+ 2y = 3 and - 2x + y = 1−2x+y=1- 2x + y = 1?

1 Answer

Explanation:

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Impact of this question

How do you solve the system of equations $x + 2 y = 3$ $x+ 2y = 3$ and $- 2 x + y = 1$ $- 2x + y = 1$ ?