How do you solve the following system: $5x+2y=20 , 3x+4y=-10$ ?

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Answer 1 · 2016-03-03T18:02:43

The solution for the system of equations is:

$color(blue)(x=50/7$

$color(blue)(y = -55/7$

$5x+2y=20$ , multiplying by $2$
$10x+color(blue)(4y)=40$ .......equation $(1)$

$3x +color(blue)(4y)=-10$ .....equation $(2)$

Solving by elimination.

Subtracting equation $(2)$ from $(1)$ results in elimination of $color(blue)(4y$

$10x+cancelcolor(blue)(4y)=40$
$-3x -cancelcolor(blue)(4y)=10$

$7x=50$

$color(blue)(x=50/7$

Finding $y$ from equation $1$

$5x+2y=20$

$2y=20 -5x$

$2y=20 -5 xx 50/7$

$2y=20 - 250/7$

$2y=140/7 - 250/7$

$2y=- 110/7$

$y=- 110/(7 xx2)$

$color(blue)(y = -55/7$

How do you solve the following system: 5x+2y=20 , 3x+4y=-10 5x+2y=20 , 3x+4y=-10 ?