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Solving system by substitution

Solving system by substitution-example-1
User LouieV
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Given the system of equations below:


\large{ \begin{cases} x + 2y = 12 \\ x = y - 12 \end{cases}}

For the second equation, x-term is isolated and can be substituted in the first equation.


\large{(y - 12) + 2y = 12}

The equation above is when we substitute x = y-12 in the first equation. Cancel the brackets.


\large{y - 12 + 2y = 12}

Add up the like term and isolate y-term.


\large{3y - 12= 12} \\ \large{3y - 12 + 12 = 12 + 12}

Add both sides by 12 to get rid of 12 from the left side to isolate y-term.


\large{3y = 24}

Divide both sides by 3 so we can finally isolate the term.


\large{ (3y)/(3) = (24)/(3) } \\ \large{ \frac{ \cancel{3}y}{ \cancel{3}} = \frac{ \cancel{24}}{ \cancel{3}} } \\ \large{y = 8}

Next, find the x-value because in system of equations - we have to answer as in an ordered pairs or coordinate point. We know y-value now but we don't know x-value yet. To find x-value, we substitute the y-value in one of two equations that are given. You can substitute in both equation but it's not necessary to substitute in both equations at one. I will choose to substitute in x = y-12.


\large{x = y - 12}

Substitute y = 8 in the equation.


\large{x = 8 - 12} \\ \large{x = - 4}

Now that we know the both values. We finally have an answer to this problem. Hence.

Answer

  • x = -4, y = 8
  • (-4,8)

The second answer is in ordered pair form. Let me know if you have any doubts!

User Ewooycom
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