How do you solve $x + y = 3$ $x + y = 3$ and $3 x + 4 y = 8$ $3x + 4y = 8$ using substitution?

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Answer 1 · 2016-12-06T21:36:46

$x = 4$ $x = 4$ and $y = - 1$ $y = -1$

Step 1) Solve the first equation for $y$ $y$ :

$x - x + y = 3 - x$ $x - x + y = 3 - x$

$y = 3 - x$ $y = 3 - x$

Step 2) Substitute $3 - x$ $3 - x$ for $y$ $y$ in the second equation and solve for $x$ $x$ :

$3 x + 4 (3 - x) = 8$ $3x + 4(3 - x) = 8$

$3 x + 12 - 4 x = 8$ $3x + 12 - 4x = 8$

$12 - x = 8$ $12 - x = 8$

$12 - 8 - x + x = 8 - 8 + x$ $12 - 8 - x + x = 8 - 8 + x$

$4 = x$ $4 = x$

$x = 4$ $x = 4$

Step 3) substitute $4$ $4$ for $x$ $x$ in the solution to the first equation in Step 1) and calculate $y$ $y$ :

$y = 3 - 4$ $y = 3 - 4$

$y = - 1$ $y = -1$

How do you solve x + y = 3 x+y=3x + y = 3 and 3x + 4y = 83x+4y=83x + 4y = 8 using substitution?