How do you solve the following system: $x + 8y = 15 , 2x + 5y= 2$ ?

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Answer 1 · 2015-12-30T11:19:42

x = $-169/11$
y = $28/11$

x + 8y = 15 --------------(1)
2x + 5y = 2 -------------(2)

Multiply equation (1) by 2, we get

2x + 16y = 30 ----------(3)

subtract equation 2 from equation 3 we get,

11y = 28

y = $28/11$

substitute in equation in 1
x + 8( $28/11$ ) = 15

x+ $224/11$ = 15

x= 15- $224/11$

x = $(55-224)/11$

x= $-169/11$

How do you solve the following system: x + 8y = 15 , 2x + 5y= 2 x + 8y = 15 , 2x + 5y= 2 ?