Substitution or elimination

[JayJay06]

Solve by using Substitution or elimination:

2x - 4y = -2 and 4x + 8y = 0

Answer:

2x - 4y -x = -2 - 2x
-4y = -2 -2x

4x + 8(-2 - 2x) = 0
4x + 16 -16x = 0
16 - 12x = 0
-12x = -16
x = -1 1/3

4(-1 1/3) + 8y = 0
-5 1/3 + 8y =0
8y = 5 1/3
y = 2/3

(- 1 1/3 , 2/3)

 

[Denis]

Solve by using Substitution or elimination:
2x - 4y = -2 and 4x + 8y = 0

Quite a few errors in your solution, JJ; were you absent
when this was covered in class?

Start by simplifying; divide 1st equation by 2, and 2nd by 4:
x - 2y = -1
x + 2y = 0

Using 1st equation:
x - 2y + 2y = 2y - 1
x = 2y - 1

Finish it :idea:

 

[soroban]

Hello, JayJay06!

Solve by using substitution or elimination: \(\displaystyle \:\begin{array}{cc}(1)\\(2)\end{array}\;\begin{array}{cc}2x\,-\,4y & = & -2 \\ 4x\,+\,8y & = & 0\end{array}\)

Substitution

Solve (2) for \(\displaystyle x:\;\;4x\,+\,8y\:=\:0\;\;\Rightarrow\;\;4x\:=\:-8y\;\;\Rightarrow\;\;x\:=\:-2y\)

Substitute into (1): \(\displaystyle \:2(-2y)\,-\,4y\:=\:-2\)

Solve for \(\displaystyle y:\;\;-4y\,-\,4y\:=\:-2\;\;\Rightarrow\;\;-8y\:=\:=2\;\;\Rightarrow\;\;\fbox{y \,=\,\frac{1}{4}}\)

Substitute into (2): \(\displaystyle \:4x\,+\,8\left(\frac{1}{4}\right)\:=\:0\;\;\Rightarrow\;\;4x\:=\;-2\;\;\Rightarrow\;\;\fbox{x\.=\.-\frac{1}{2}}\)


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Elimination

\(\displaystyle \begin{array}{cc}\text{Multiply (1) by 2:}\\\text{Add (2):}\end{array}\;\;
\begin{array}{cc}4x\,-\,8y & = & -4 \\ 4x\,+\,8y & = & 0\end{array}\)

. . and we have: \(\displaystyle \:8x\:=\:-4\;\;\Rightarrow\;\;\fbox{x\:=\:-\frac{1}{2}}\)

Substitute into (2): \(\displaystyle \:4\left(-\frac{1}{2}\right)\,+\,8y\:=\:0\;\;\Rightarrow\;\;8y\:=\:2\;\;\Rightarrow\;\;\fbox{y\:=\:\frac{1}{4}}\)


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Check

Substitute into (1): \(\displaystyle \:2\left(-\frac{1}{2}\right)\,-\,4\left(\frac{1}{4}\right)\:=\:-1\,-\,1\:=\:-2\) . . . yes!

Substitute into (2): \(\displaystyle \:4\left(-\frac{1}{2}\right)\,+\,8\left(\frac{1}{4}\right) \:=\:-2\,+\,2\:=\:0\) . . . yes!