# How do you solve the system of equations 3x - 10y = - 20 and 6x - 8y = 20?

Dec 15, 2016

$\left\{x = 10 , y = 5\right]$

#### Explanation:

$3 x - 10 y = - 20 \text{ } \left(1\right)$

$6 x - 8 y = 20 \text{ } \left(2\right)$

$\text{you can multiply the both side of equation (1) by 2}$

$\textcolor{red}{2} \left(3 x - 10 y\right) = \textcolor{red}{2} \left(- 20\right)$

$\text{rearrange }$

$6 x - 20 y = - 40 \text{ } \left(3\right)$

$\text{now subtract (3) from (2)}$

$6 x - 8 y - \left(6 x - 20 y\right) = 20 - \left(- 40\right)$

$\text{rearrange the equation}$

$\cancel{6 x} - 8 y - \cancel{6 x} + 20 y = 20 + 40$

$12 y = 60$

$\text{divide both side by 12}$

$\frac{\cancel{12} y}{\cancel{12}} = \frac{60}{12} \text{ ; } y = \frac{60}{12}$

$\textcolor{g r e e n}{y = 5}$

$\text{now use (1) or (2)}$

$3 x - 10 \cdot \textcolor{g r e e n}{5} = - 20 \text{ } \left(1\right)$

$3 x - 50 = - 20$

$3 x = - 20 + 50$

$3 x = 30$

$\text{divide both side of equation by 3}$

$\frac{\cancel{3} x}{\cancel{3}} = \frac{30}{3}$

$x = \frac{30}{3}$

$x = 10$

Dec 15, 2016

Different approach

$x = 10 \text{; } x = 5$ as solutions to this system of equations

#### Explanation:

Given:

$3 x - 10 y = - 20. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$
$6 x - 8 y = 20. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(2\right)$
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Method: use one equation to express what y is 'worth' (dependant variable). Then substitute for y in the other equation so that you can solve for $x$.

Consider Equation(1)

$10 y = 3 x + 20$

$y = \frac{3}{10} x + 2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left({1}_{a}\right)$

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

$6 x - 8 y = 20 \text{ "->" } 6 x - 8 \left(\frac{3}{10} x + 2\right) = 20$

$\text{ } 6 x - \frac{24}{10} x - 16 = 20$

$\text{ } \frac{36}{10} x = 36$

$\text{ } \frac{1}{10} x = 1$

$\text{ } \textcolor{p u r p \le}{x = 10}$
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I chose equation(1) as the number would be easier to work out.

Substitute $x = 10$ into equation(1) giving

$3 x - 10 y = - 20 \text{ "->" } 3 \left(10\right) - 10 y = - 20$

$\text{ } - 10 y = - 50$

$\text{ } \textcolor{p u r p \le}{y = + 5}$
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