How do solve the following linear system?: $4x+3y=1 , 11x + 3y + 7 = 0$ ?

Question

1535 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-05T06:37:35

$x=-8/7=-1 1/7$

$y=13/7=1 6/7$

Given: $4x+3y=1$ ; $11x+3y+7=0$

From $4x+3y=1$ we can subtract $4x$ from both sides

to get $3y=1-4x$ which we can substitute into the other

equation for $3y$ :

$11x+3y+7=0$

$11x+1-4x+7=0$

$7x=-8$

$x=-8/7$ answer x

Substitute value for $x$ into $4x+3y=1$ to find $y$ :

$4(-8/7)+3y=1$

$3y=1+32/7$

$3y=39/7$

$y=13/7$ answer y

To check, substitute into the $given$ equation:

$11x+3y+7=0$

$11(-8/7)+3(13/7)+7=0$

$(-88/7)+(39/7)+(49/7)=0$

$(-88)+(39)+(49)=0$

$-88+88=0$

How do solve the following linear system?: 4x+3y=1 , 11x + 3y + 7 = 0 4x+3y=1 , 11x + 3y + 7 = 0 ?