How do I use substitution to find the solution of the system of equations $y = \frac{1}{3} x + \frac{7}{3}$ $y=1/3x+7/3$ and $y = - \frac{5}{4} x + \frac{11}{4}$ $y=-5/4x+11/4$ ?

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How do I use substitution to find the solution of the system of equations $y = \frac{1}{3} x + \frac{7}{3}$ $y=1/3x+7/3$ and $y = - \frac{5}{4} x + \frac{11}{4}$ $y=-5/4x+11/4$ ?

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Answer 1 · 2014-09-26T23:56:27

In this problem we have 2 different expressions that represent the value of $y$ $y$ . Because they both represent $y$ $y$ we can set them equal to each other.

$(\frac{1}{3}) x + \frac{7}{3} = - (\frac{5}{4}) x + \frac{11}{4}$ $(1/3)x+7/3=-(5/4)x+11/4$

Because all of the terms share a common multiple of 12 I will multiply each term by 12 to eliminate the fractions.

$12 [(\frac{1}{3}) x + \frac{7}{3} = - (\frac{5}{4}) x + \frac{11}{4}]$ $12[(1/3)x+7/3=-(5/4)x+11/4]$

$[(\frac{12}{3}) x + \frac{84}{3} = - (\frac{60}{4}) x + \frac{132}{4}]$ $[(12/3)x+84/3=-(60/4)x+132/4]$

$4 x + 28 = - 15 x + 33$ $4x+28=-15x+33$

$19 x = 5$ $19x=5$

$x = \frac{5}{19}$ $x=5/19$

Substitute in this newly found $x$ $x$ -value into one of the original equations.

$y = (\frac{1}{3}) x + \frac{7}{3}$ $y=(1/3)x+7/3$

$y = (\frac{1}{3}) \cdot (\frac{5}{19}) + \frac{7}{3}$ $y=(1/3)*(5/19)+7/3$

$y = (\frac{5}{57}) + \frac{7}{3} \cdot \frac{19}{19}$ $y=(5/57)+7/3*19/19$

$y = \frac{5}{57} + \frac{133}{57}$ $y=5/57+133/57$

$y = \frac{138}{57}$ $y=138/57$

$y = \frac{46}{19}$ $y=46/19$

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How do I use substitution to find the solution of the system of equations $y = \frac{1}{3} x + \frac{7}{3}$ $y=1/3x+7/3$ and $y = - \frac{5}{4} x + \frac{11}{4}$ $y=-5/4x+11/4$ ?

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How do I use substitution to find the solution of the system of equations $y = \frac{1}{3} x + \frac{7}{3}$ $y=1/3x+7/3$ and $y = - \frac{5}{4} x + \frac{11}{4}$ $y=-5/4x+11/4$ ?