How do you solve the following system:  8x-7y=-39, 7x − 5y = −11 ?

Nov 5, 2017

$x = \frac{118}{9} , y = \frac{185}{9}$

Explanation:

$8 x - 7 y = - 39$...(1)
$7 x - 5 y = - 11$...(2)

(1)*5:
$40 x - 35 y = - 195$...(3)

(2)*7:
$49 x - 35 y = - 77$...(4)

(4)-(3):
$9 x = 118$
$x = \frac{118}{9}$

Sub x=118/9 into (1)
$\frac{944}{9} - 7 y = - 39$
$- 7 y = - \frac{1295}{9}$
$y = \frac{185}{9}$

Nov 5, 2017

I have solved $\frac{1}{2}$ of it left the remainder for you to do.
Check your solution with that in the graph.

Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

Example of the principle used. When I realised this (long time ago) it opened the door for me to a whole new way of thinking.

Suppose we have $\textcolor{w h i t e}{\text{d}} \textcolor{g r e e n}{3 x - 6 = 5}$

Then it is also true that

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddd}} \textcolor{red}{2 \times} \left(3 x - 6\right) = \textcolor{red}{2 \times} \left(5\right)}$
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$\textcolor{b l u e}{\text{Answering the question}}$

Given:
$8 x - 7 y = - 39 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$
$7 x - 5 y = - 11 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$

If we change one or both of these then subtract it will eliminate one of the unknown. Then we have 1 equation with 1 unknown and thus solvable

Lets 'get rid' of the $y ' s$

Multiply $E q n \left(1\right)$ by 5 and $E q n \left(2\right)$ by 7

color(white)("d")40x-35y=-195" ".........Equation(1_a)
$\textcolor{w h i t e}{\text{d")ul(49x-35y=-77" "............Equation(2_a)larr" Subtract}}$
$- 9 x + \textcolor{w h i t e}{\text{d}} 0 y = - 118$

Divide both side by -9

$x = + \frac{118}{9}$

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I will let you determine $y$. Just substitute this value into one of the equation.