# Substitution equation Problem

• November 16th 2006, 05:51 AM
mathnoobie25
Substitution equation Problem
using the substitution method find the solution to each system of equation...
1.
2x+3y=7
6x-y=1

idk if i got it right but i got y=2.5 and x=1/4 it seems wrong though...
• November 16th 2006, 06:39 AM
Soroban
Hello, mathnoobie25!

Quote:

Using the substitution method, solve: $\begin{array}{cc}(1)\\(2)\end{array}\begin{array}{ cc}2x + 3y \:=\:7 \\ 6x - y\:=\:1\end{array}$

idk if i got it right but i got y=2.5 and x=1/4 it seems wrong though.
. .
Don't you know how to check your answers?

Solve one of the equations for one of its variables.
. . (Try to choose the easiest one.) *

Solve equation (2) for $y:\;\;y \:=\:6x - 1$ (3)

Substitute into (1): . $2x + 3(6x - 1) \:=\:7$

Solve for $x:\;\;2x + 18x - 3\:=\:7\quad\Rightarrow\quad 20x \:=\:10\quad\Rightarrow\quad\boxed{ x \,= \,\frac{1}{2}}$

Substitute into (3): . $y \:=\:6\left(\frac{1}{2}\right) - 1\quad\Rightarrow\quad\boxed{ y\,=\,2}$

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Check

Substitute your answers into the original equations
. . and see if you get true statements.

Substitute $\left(\frac{1}{2},\,2\right)$ into (1).
. . $2\left(\frac{1}{2}\right) + 3(2)\:=\:7$ . . . Is this true?
We have: . $1 + 6 \,=\,7$ . . . Yes!

Substitute $\left(\frac{1}{2},\;2\right)$ into (2).
. . $6\left(\frac{1}{2}\right) - 2 \,=\,1$ . . . Is this true?
We have: . $3 - 2\,=\,1$ . . . Yes!

Therefore, our answers are correct.

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*
We could, say, solve (1) for $x:\;\;2x\:=\:7 - 3y\quad\Rightarrow\quad x \:=\:\frac{7}{2} - \frac{3}{2}y$

. . But why introduce fractions when we can avoid them?