Systems of Equations Question 2

What is the value of $x-y$, given the following system of equations?

$$ 7x+3y=8$$ $$ 6x-3y=5$$

$-\dfrac{4}{3}$

$\dfrac{1}{3}$

$\dfrac{2}{3}$

$1$

Approach 1

Because the equations are arranged so that corresponding terms are matching, we can use elimination and add each corresponding term.

$$ 7x+3y=8$$ $$ 6x-3y=5$$ $$ \begin{array}{c c c} &7x & 3y & 8 \\ +&6x & -3y & 5 \\ \hline \end{array} $$ $$13x=13$$ $$x=1$$

Substituting $x$ into either equation,

$$ 7x+3y=8$$ $$7(1)+3y=8$$ $$7+3y=8$$ $$3y=1$$ $$y=\frac{1}{3}$$

Finally,

$$x-y = 1-\frac{1}{3}$$ $$ = \boxed{\frac{2}{3}}$$

Approach 2

Although elimination is efficient, we can also use substitution to eliminate a variable. Note that the $3y$ term is present in both equations.

$$ 7x+\colorbox{yellow}{$3y$}=8$$ $$ 6x-\colorbox{yellow}{$3y$}=5$$

Solving for $3y$ in the first equation,

$$7x+3y=8$$ $$3y=8-7x$$

Substituting this value into the second equation,

$$ 6x-3y=5$$ $$6x-(8-7x)=5$$ $$6x-8+7x=5$$ $$13x=13$$ $$x=1$$

Substituting $x$ into either equation,

$$ 7x+3y=8$$ $$7(1)+3y=8$$ $$7+3y=8$$ $$3y=1$$

Finally,

$$x-y = 1-\frac{1}{3}$$ $$ = \boxed{\frac{2}{3}}$$