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Thread: substitution

  1. #1
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    Exclamation substitution

    alright i know how you substitute when they give you something easy like
    6x+7=5
    y=5+3
    because there's already a 'y='


    But when you have something like
    9x-7y=5
    5x+3y=-1 i don't understand you you can make the y= or x=

    could someone please help me?
    thank you
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  2. #2
    Senior Member tukeywilliams's Avatar
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    So you have:

    $\displaystyle 9x-7y = 5 $
    $\displaystyle 5x + 3y = -1 $

    So first decide what variable you want to solve for in terms of the other variable. I want to solve for $\displaystyle x $ in terms of $\displaystyle y $. So I look at the first equation:

    $\displaystyle 9x-7y = 5 $
    $\displaystyle 9x = 5+7y $

    $\displaystyle x = \frac{5+7y}{9} $

    So now I have $\displaystyle x $ in terms of $\displaystyle y $.

    Now go to the second equation and substitute in this expression.

    $\displaystyle 5\left(\frac{5+7y}{9}\right ) + 3y = -1 $. You can then solve for $\displaystyle y $. After solving for $\displaystyle y $ plug this value back into the first equation and solve for $\displaystyle x $.

    So you get $\displaystyle x = \; \text{something}, \; y = \; \text{something} $
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by ncbabe View Post
    ...
    But when you have something like
    9x-7y=5
    5x+3y=-1 i don't understand you you can make the y= or x=
    ...
    Hello,

    1. You have to determine which variable should be substituted. In this example I will substitute the y of the first equation by the y of the second equation.

    2. Thus we need the y of the second equation in explicite(?) form:

    $\displaystyle 5x+3y=-1 \Longleftrightarrow 3y=-5x-1 \Longleftrightarrow y=-\frac{5}{3}x-\frac{1}{3}$

    3. Now substitute the y of the first equation by the term for y:

    $\displaystyle 9x-7 \cdot \left(-\frac{5}{3}x-\frac{1}{3} \right)=5 \Longleftrightarrow 9x+\frac{35}{3}x+\frac{7}{3}=5$

    4. Collect like terms: x-values at the LHS and all constant values at the RHS:
    $\displaystyle \frac{62}{3}x=\frac{8}{3}$

    5. Divide by the coefficient of x:

    $\displaystyle x=\frac{\frac{8}{3}}{\frac{62}{3}} = \frac{4}{31}$

    EDIT:
    6. Claculate the y-value by substituting the x-value into the equation: $\displaystyle y=-\frac{5}{3}x-\frac{1}{3}$

    $\displaystyle y=-\frac{5}{3}\cdot \frac{4}{31}-\frac{1}{3} = -\frac{17}{31}$
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