Exercise 10 Page 419

To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the $x$-terms or the $y$-terms must cancel each other out. In its current state, this will not happen. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply Equation (II) by $2,$ the $y$-terms will have opposite coefficients. We can see that the $y$-terms will eliminate each other if we add Equation (I) to Equation (II).

$\{\begin{array}{l}x-4y=\text{-}1\\ 10x+4y=34\end{array}$

SysEqnAdd

$\text{(II): }$ Add $\text{(I)}$

$\{\begin{array}{l}x-4y=\text{-}1\\ 10x+4y+(x-4y)=34+(\text{-}1)\end{array}$

▼

$\text{(II): }$Solve for $x$

RemovePar

$\text{(II): }$ Remove parentheses

$\{\begin{array}{l}x-4y=\text{-}1\\ 10x+4y+x-4y=34-1\end{array}$

AddSubTerms

$\text{(II): }$ Add and subtract terms

$\{\begin{array}{l}x-4y=\text{-}1\\ 11x=33\end{array}$

DivEqn

$\text{(II): }$ $\text{LHS}/11=\text{RHS}/11$

$\{\begin{array}{l}x-4y=\text{-}1\\ x=3\end{array}$

Now we can solve for $y$ by substituting the value of $x$ into Equation (I) and simplifying.

$\{\begin{array}{l}x-4y=\text{-}1\\ x=3\end{array}$

Substitute

$\text{(I): }$ $x=3$

$\{\begin{array}{l}3-4y=\text{-}1\\ x=3\end{array}$

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$\text{(I): }$Solve for $y$

SubEqn

$\text{(I): }$ $\text{LHS}-3=\text{RHS}-3$

$\{\begin{array}{l}\text{-}4y=\text{-}4\\ x=3\end{array}$

DivEqn

$\text{(I): }$ $\text{LHS}/(\text{-}4)=\text{RHS}/(\text{-}4)$

$\{\begin{array}{l}y=1\\ x=3\end{array}$

The solution, or intersection point, of the system of equations is $(3,1).$ We will check this solution by graphing. To do this we will find $x\text{-}$ and $y\text{-intercepts}$ of both lines.

Information	$x-4y=\text{-}1$	$5x+2y=17$
$y=0$	$x-4(0)=\text{-}1$	$5x+2(0)=17$
Simplify	$x=\text{-}1$	$x=3\frac{2}{5}$
$x\text{-}$intercept	$\left(\text{-}1,0\right)$	$\left(3\frac{2}{5},0\right)$
$x=0$	$0-4y=\text{-}1$	$5(0)+2y=17$
Simplify	$y=\frac{1}{4}$	$y=8\frac{1}{2}$
$y\text{-}$intercept	$\left(0,\frac{1}{4}\right)$	$\left(0,8\frac{1}{2}\right)$

Now, we can plot these points and connect them.

The solution to the system is the point where the lines intersect. The two lines appear to intersect at the point $(3,1).$ Since the point of intersection is the same as our solution, we know that our answer is correct.