How do solve the following linear system?: $x + 4 y = 3, 6 x = 29 - 8 y$ $x+4y=3 , 6x=29-8y$ ?

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Answer 1 · 2016-01-07T17:06:39

The solution for the system of equations is

$x = \frac{23}{4}, y = - \frac{11}{16}$ $color(blue)(x=23/4 , y=-11/16$

$x + 4 y = 3$ $x+4y=3$ , multiplying by $6$ $6$
$6 x + 24 y = 18$ $color(blue)(6x)+24y=18$ .....equation $(1)$ $(1)$

$6 x = 29 - 8 y$ $6x=29-8y$

$6 x + 8 y = 29$ $6x+8y=29$ .......equation $(2)$ $(2)$

Solving by elimination.
Subtracting equation $2$ $2$ from $1$ $1$

$cancelcolor(blue)(6x)+24y=18$
$-cancelcolor(blue)(6x)-8y=-29$

$16y=-11$

$color(blue)(y=(-11)/16$

Finding $x$ from equation $1$
$x+4y=3$

$x=3-4y$
$x=3-4 xx(-11/16)$

$x=3+(44/16)$

$x=48/16+(44/16)$

$x=(92/16)$

$color(blue)(x=23/4$

How do solve the following linear system?: x+4y=3 , 6x=29-8y x+4y=3,6x=29−8y x+4y=3 , 6x=29-8y ?