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How do solve the following linear system?: $x + 2 y = 1, 4 x + 16 y = 2$ $x+2y=1 , 4 x+ 16y = 2$ ?

1 Answer

Dec 8, 2015

$(x, y) = (\frac{3}{2}, - \frac{1}{4})$ $(x,y) = (3/2, -1/4)$

Explanation:

Given
[1] $XXX x + 2 y = 1$ $color(white)("XXX")x+2y=1$
[2] $XXX 4 x + 16 y = 2$ $color(white)("XXX")4x+16y=2$

Multiply [1] by $4$ $4$ to generate the same $x$ $x$ coefficient as in [2]
[3] $XXX 4 x + 8 y = 4$ $color(white)("XXX")4x+8y=4$

Subtract [3] from [2]
[4] $XXX 8 y = - 2$ $color(white)("XXX")8y=-2$

Divide both sides of [4] by $8$ $8$
[5] $XXX y = - \frac{1}{4}$ $color(white)("XXX")y = -1/4$

Substitute $(- \frac{1}{4})$ $(-1/4)$ for $y$ $y$ in [2]
[6] $XXX 4 x + 16 \cdot (- \frac{1}{4}) = 2$ $color(white)("XXX")4x+16*(-1/4)=2$

Simplify:
[7] $XXX 4 x - 4 = 2$ $color(white)("XXX")4x-4 = 2$

Add $4$ $4$ to both sides
[8] $XXX 4 x = 6$ $color(white)("XXX")4x = 6$

Divide both sides of [8] by $4$ $4$
[9] $XXX x = \frac{3}{2}$ $color(white)("XXX")x=3/2$

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