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Math Help - solving for substitution

  1. #1
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    solving for substitution

    having trouble understanding substitution this is the problem :

    Solve for the system of equations by substitution

    5x-7=-y
    2x-y=0

    This is another one..

    6x+7y=1
    x=55-9y
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  2. #2
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    5x-7=-y
    2x-y=0.

    Substituting -y into the second equation gives

    2x+5x-7=0

    7x-7=0

    7x = 7

    x = 1.


    You know -y = 5x-7 so

    -y = 5\cdot 1 - 7

    -y = 5 - 7

    -y = -2

    y = 2.


    Therefore (x, y) = (1, 2) is the solution.
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  3. #3
    Senior Member eumyang's Avatar
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    5x-7=-y
    2x-y=0

    First, solve one of the equations for one of the variables. It really doesn't matter which equation or which variable. You probably will want to choose the way that's the easiest. I'm going to start by solving the 2nd equation for y:

    \begin{aligned}<br />
2x - y &= 0 \\<br />
2x &= y \\<br />
\end{aligned}

    Now, substitute this into the 1st equation wherever you see a y, and then solve for x:
    \begin{aligned}<br />
5x - 7 &= -y  \\<br />
5x - 7 &= -2x \\<br />
-7 &= -7x \\<br />
x &= 1<br />
    \end{aligned}

    Now, plug this into the altered version of the 2nd equation:
    \begin{aligned}<br />
2x &= y \\<br />
2(1) &= y \\<br />
y &= 2<br />
\end{aligned}

    The answer is (1, 2). You want to try your other example now?

    EDIT: Too slow!
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  4. #4
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    thank you! does anyone know solving for elimination?

    4x=15+3y
    -6/5x+y=-17/5
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  5. #5
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    I can't really read that...

    Is it

    4x=15+3y
    -\frac{6}{5x}+y = -\frac{17}{5}?
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  6. #6
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    yes the 6/5x , the x is for both the 6 and 5 not just the 5.. not even sure if that matters but you have it right.
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  7. #7
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    Obviously it's not right if it's meant to be

    4x=15+3y
    -\frac{6}{5}x+y = -\frac{17}{5}

    since \frac{6}{5}x = \frac{6x}{5}, not \frac{6}{5x}...


    Anyway...

    4x=15+3y
    -\frac{6}{5}x+y = -\frac{17}{5}

    Multiply the second equation by 3...

    4x=15+3y
    -\frac{18}{5}x+3y=-\frac{51}{5}

    4x-3y=15
    -\frac{18}{5}x+3y=-\frac{51}{5}.


    Now add the equations together

    (4x-3y) + \left(-\frac{18}{5}x+3y\right) = 15-\frac{51}{5}

    \frac{2}{5}x=\frac{24}{5}

    2x = 24

    x = 12.


    Substituting into the first equation

    4x=15+3y

    4\cdot 12 = 15 + 3y

    48 = 15 + 3y

    33 = 3y

    y = 11.


    So (x, y) = (12, 11) is the solution.
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  8. #8
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    Quote Originally Posted by eumyang View Post
    5x-7=-y
    2x-y=0

    First, solve one of the equations for one of the variables. It really doesn't matter which equation or which variable. You probably will want to choose the way that's the easiest. I'm going to start by solving the 2nd equation for y:

    \begin{aligned}<br />
2x - y &= 0 \\<br />
2x &= y \\<br />
\end{aligned}

    Now, substitute this into the 1st equation wherever you see a y, and then solve for x:
    \begin{aligned}<br />
5x - 7 &= -y  \\<br />
5x - 7 &= -2x \\<br />
-7 &= -7x \\<br />
x &= 1<br />
    \end{aligned}

    Now, plug this into the altered version of the 2nd equation:
    \begin{aligned}<br />
2x &= y \\<br />
2(1) &= y \\<br />
y &= 2<br />
\end{aligned}

    The answer is (1, 2). You want to try your other example now?

    EDIT: Too slow!
    yeah if you can help me out with the other example that would be great.. this is one part of math that i do not like haha
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  9. #9
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    double posted by accident sorry
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  10. #10
    Senior Member eumyang's Avatar
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    OP: Prove It's point is that you have to be careful with notation. Ideally, you should learn LaTeX so that it's not ambiguous when you type \frac{6}{5}x. 6/5x is really read as \frac{6}{5x}. If you are not using LaTeX and you want to indicate the fraction 6/5 times x, use parentheses: (6/5)x.
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