## Precalculus (6th Edition) Blitzer

Let us consider the following example of the substitution method to see which method is easier: $y=3-3x$ ( $1$ ) $3x+4y=6$ ( $2$ ) Put the value of y from the first equation in the second equation as follows: \begin{align} & 3x+4y=6 \\ & 3x+4\cdot \left( 3-3x \right)=6 \\ & 3x+12-12x=6 \end{align} And subtract like terms as follows: $12-9x=6$ Add $-12$ to both sides as follows: \begin{align} & 12-9x-12=6-12 \\ & -9x=-6 \\ & x=\frac{2}{3} \end{align} Then, put the value of x in the first equation as follows: \begin{align} & y=3-3x \\ & y=3-3\cdot \frac{2}{3} \\ & y=1 \\ \end{align} Now, consider the following example of the addition method to compare between the addition method and the substitution method: $3x+5y=-2$ ( $1$ ) $2x+3y=0$ ( $2$ ) And multiply by $-2$ on both sides of the first equation as follows: \begin{align} & 3x+5y=-2 \\ & -2\left( 3x+5y \right)=-2\cdot -2 \end{align} $-6x-10y=4$ (a) Also, multiply by $3$ on both sides of the second equation as follows: \begin{align} & 2x+3y=0 \\ & 3\left( 2x+3y \right)=0\cdot 3 \end{align} $6x+9y=0$ (b) Then by the addition method just add (a) and (b) as follows: \begin{align} & -6x-10y+6x+9y=4+0 \\ & -y=4 \\ & y=-4 \end{align} Put the value of $y$ in (b) as follows: \begin{align} & 6x+9\cdot -4=0 \\ & 6x-36=0 \end{align} Add to both sides $36$ as follows: \begin{align} & 6x-36+36=0+36 \\ & 6x=36 \end{align} Multiply both sides by $\frac{1}{6}$ as follows: \begin{align} & \frac{1}{6}\cdot 6x=\frac{1}{6}\cdot 36 \\ & x=6 \\ \end{align} We see that it depends on the question which method is easier to follow; the steps required in both methods are the same.