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

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Answer 1 · 2017-05-05T06:37:35

$x = - \frac{8}{7} = - 1 \frac{1}{7}$ $x=-8/7=-1 1/7$

$y = \frac{13}{7} = 1 \frac{6}{7}$ $y=13/7=1 6/7$

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

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

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

equation for $3 y$ $3y$ :

$11 x + 3 y + 7 = 0$ $11x+3y+7=0$

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

$7 x = - 8$ $7x=-8$

$x = - \frac{8}{7}$ $x=-8/7$ answer x

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

$4 (- \frac{8}{7}) + 3 y = 1$ $4(-8/7)+3y=1$

$3 y = 1 + \frac{32}{7}$ $3y=1+32/7$

$3 y = \frac{39}{7}$ $3y=39/7$

$y = \frac{13}{7}$ $y=13/7$ answer y

To check, substitute into the $g i v e n$ $given$ equation:

$11 x + 3 y + 7 = 0$ $11x+3y+7=0$

$11 (- \frac{8}{7}) + 3 (\frac{13}{7}) + 7 = 0$ $11(-8/7)+3(13/7)+7=0$

$(- \frac{88}{7}) + (\frac{39}{7}) + (\frac{49}{7}) = 0$ $(-88/7)+(39/7)+(49/7)=0$

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

$- 88 + 88 = 0$ $-88+88=0$

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