How do you solve the following system?: $5 x + 2 y = 1, - 8 x + 3 y = 12$ $5x + 2y =1 , -8x+3y=12$

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How do you solve the following system?: $5 x + 2 y = 1, - 8 x + 3 y = 12$ $5x + 2y =1 , -8x+3y=12$

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Answer 1 · 2017-06-23T17:58:28

$(- \frac{21}{31}, \frac{68}{31})$ $(-21/31, 68/31)$

Explanation:

$5 x + 2 y = 1$ $5x+2y=1$
$- 8 x + 3 y = 12$ $-8x+3y=12$

The best way to solve this system of equations is by using the elimination method. Basically, we need to eliminate a variable by adding the two equations together.

However, to completely cancel out a variable, such as $x$ $x$ , they need to have the same coefficient but different signs (positive and negative).

$8 (5 x + 2 y) = (1) 8$ $8(5x+2y)=(1)8$
$5 (- 8 x + 3 y) = (12) 5$ $5(-8x+3y)=(12)5$

$40 x + 16 y = 8$ $40x+16y=8$
$- 40 x + 15 y = 60$ $-40x+15y=60$

By multiplying the equations, we can now safely eliminate $x$ $x$ from the system by adding the two equations together.

$31 y = 68$ $31y=68$

$y = \frac{68}{31}$ $y=68/31$

Now, you have to plug $y$ $y$ back into one of the equations to get $x$ $x$ .

$5 x + 2 (\frac{68}{31}) = 1$ $5x+2(68/31)=1$

$5 x + \frac{136}{31} = 1$ $5x+136/31=1$

$5 x = 1 - \frac{136}{31}$ $5x=1-136/31$

$5 x = - \frac{105}{31}$ $5x=-105/31$

$x = - \frac{105}{31} \cdot \frac{1}{5}$ $x=-105/31 * 1/5$

$x = - \frac{21}{31}$ $x=-21/31$

Here's your answer:

$(- \frac{21}{31}, \frac{68}{31})$ $(-21/31, 68/31)$

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How do you solve the following system?: $5 x + 2 y = 1, - 8 x + 3 y = 12$ $5x + 2y =1 , -8x+3y=12$

1 Answer

Explanation:

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How do you solve the following system?: $5 x + 2 y = 1, - 8 x + 3 y = 12$ $5x + 2y =1 , -8x+3y=12$