# How do you solve the system 3x-5y=7 and 2x+5y=13?

Feb 21, 2016

Point of intersection as at$\text{ "(x,y)" "->" } \left(4 , 1\right)$

#### Explanation: Always look for the easiest way, if you can spot it that is! Sometimes you have no choice but to just jump in and do it the hard way. Not so in this case!

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$\textcolor{b l u e}{\text{Observation 1}}$

Notice that both equations are in the same order

$\textcolor{b l u e}{\text{Observation 2 }}$

Notice that the coefficients of $y$ are of the same magnitude but of opposite sign. This makes life so much easier!

$\textcolor{b l u e}{\text{Objective}}$
To end up with only one variable and a pile of numeric values. This enables the determination of the variables value.

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$\textcolor{b l u e}{\text{Solving for } x}$

Write as:
$3 x - 5 y = 7 \text{ }$.....................................(1)
$2 x + 5 y = 13 \text{ }$..................................(2)

Add equation (1) to equation (2) to 'get rid of' $y$

$5 x + 0 = 20$

Divide both sides by 5 so that you end up with $x$ on its own.

$\frac{5}{5} \times x = \frac{20}{5}$

But $\frac{5}{5} = 1$ giving:

$\textcolor{m a \ge n t a}{x = 4}$

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$\textcolor{b l u e}{\text{Solving for } y}$

Substitute $x = 4$ into either equation (1) or (2)

Selecting equation (2) as $y$ is positive, thus less work!

$2 \left(\textcolor{m a \ge n t a}{4}\right) + 5 y = 13 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$

Subtract 8 from both sides

$5 y = 5$

Divide both sides by 5

$\textcolor{m a \ge n t a}{y = 1}$

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Point of intersection as at$\text{ "(x,y)" "->" } \left(4 , 1\right)$