How do you solve the system of equations $y = 2 x + 8$ $y= 2x + 8$ and $3 x + 5 y = 1$ $3x + 5y = 1$ ?

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Answer 1 · 2018-06-26T10:51:42

By arranging the equations

Arrange the equations (multiply the 1st by 3 and the 2nd by 2):

$3 y - 6 x = 24$ $3y - 6x = 24$
$6 x + 10 y = 2$ $6x + 10y = 2$

Combine these:
$13 y = 26$ $13 y = 26$

$y = \frac{26}{13} = 2$ $y= 26/13 = 2$

Put this in the first original equation:
$2 x = y - 8$ $2x = y-8$

$2 x = 2 - 8$ $2x = 2-8$

$2 x = - 6$ $2x = -6$

$x = - 3$ $x=-3$

Answer 2 · 2018-06-26T10:54:24

$x = - 3$ $x=-3$ and $y = 2$ $y=2$

After putting first equation into second,

$3 x + 5 \cdot (2 x + 8) = 1$ $3x+5*(2x+8)=1$

$13 x + 40 = 1$ $13x+40=1$

$13 x = - 39$ $13x=-39$ , so $x = - 3$ $x=-3$

Thus, $y = 2 \cdot (- 3) + 8 = 2$ $y=2*(-3)+8=2$

How do you solve the system of equations y= 2x + 8y=2x+8y= 2x + 8 and 3x + 5y = 13x+5y=13x + 5y = 1?