How do you solve $y = 3 - \frac{1}{2} x, 3 x + 4 y = 1$ $y = 3 - 1/2 x, 3x + 4y = 1$ ?

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Answer 1 · 2016-06-26T16:59:02

$x = - 11, y = 8 \frac{1}{2}$ $x=-11, y = 8 1/2$

Substitution seems quite a good method to choose, because $y$ $y$ is given as the subject in one of the equations.

$y = (3 - \frac{1}{2} x)$ $y = (3-1/2x)$ can be substituted into $3 x + 4 y = 1$ $3x+4y =1$

ie. Use $(3 - \frac{1}{2} x)$ $(3-1/2x)$ in the place of $y$ $y$ in the second equation:

$3 x + 4 (3 - \frac{1}{2} x) = 1$ $3x+4(3-1/2x) =1" "$ now there are only x-terms.

$3 x + 12 - 2 x = 1$ $3x +12 -2x = 1$

$x = - 11$ $x = -11$

Now that we have the value for $x$ $x$ , use it to find the value of $y$ $y$ .

$y = (3 - \frac{1}{2} x) \Rightarrow = 3 - \frac{1}{2} (- 11)$ $y = (3-1/2x) rArr = 3-1/2(-11)$

$y = 3 + 5 \frac{1}{2} \Rightarrow y = 8 \frac{1}{2}$ $y = 3+5 1/2 rArr y = 8 1/2$

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