Find the solution to the following system of equations: [latex]3x-4y=8[/latex] and [latex]5x+4y=-24.[/latex]

First, line them up over top of each other, since they will be added or subtracted. Notice that, when added, the [latex]-4y[/latex] and [latex]+4y[/latex] cancel each other out:

[latex]\begin{array}{rrrrrl} &3x&-&4y&=&\phantom{-0}8 \\ +&5x&+&4y&=&-24 \\ \hline &&&\dfrac{8x}{8}&=&\dfrac{-16}{8} \\ \\ &&&x&=&-2 \end{array}[/latex]

It is now known that these equations intersect at the value where [latex]x = -2.[/latex] Now choose one of the two original equations (generally, the simplest to work with) and substitute [latex]x = -2[/latex] to find [latex]y[/latex]:

[latex]\begin{array}{rrrrr} 3(-2)&-&4y&=&8 \\ -6&-&4y&=&8 \\ +6&&&&+6 \\ \hline &&\dfrac{-4y}{-4}&=&\dfrac{14}{-4} \\ \\ &&y&=&\dfrac{14}{-4} \\ \\ &&y&=&-\dfrac{7}{2} \end{array}[/latex]

The intersection point of these two linear equations is [latex]x = -2[/latex] and [latex]y =-\dfrac{1}{2}[/latex], or at the coordinate [latex]\left(2, -\dfrac{1}{2}\right).[/latex]

Find the solution to the following system of equations: [latex]-6x+5y=22[/latex] and [latex]2x+3y=2.[/latex]

First, line up the equations and choose the variable that shall be eliminated:

[latex]\left\{ \begin{array}{rrrrr} -6x&+&5y&=&22 \\ 2x&+&3y&=&2 \end{array}\right.[/latex]

The [latex]x[/latex] variable could be eliminated if the bottom [latex]2x[/latex] were [latex]6x.[/latex] For this to happen, the entire bottom equation would have to be multiplied by 3:

[latex]\begin{array}{rrrrrrr} &(2x&+&3y&=&2)&(3) \\ \\ &-6x&+&5y&=&22& \\ +&6x&+&9y&=&6& \\ \hline &&&\dfrac{14y}{14}&=&\dfrac{28}{14}& \\ \\ &&&y&=&2& \end{array}[/latex]

It is now known that these equations intersect at the value where [latex]y = 2.[/latex] Now choose one of the two original equations (the simplest looks to be [latex]2x + 3y = 2[/latex]) and substitute [latex]y = 2[/latex] to find [latex]x[/latex]:

[latex]\begin{array}{rrcrr} 2x&+&3(2)&=&2 \\ 2x&+&6&=&2 \\ &-&6&&-6 \\ \hline &&\dfrac{2x}{2}&=&\dfrac{-4}{2} \\ \\ &&x&=&-2 \end{array}[/latex]

The intersection point of these two linear equations is [latex]x = -2[/latex] and [latex]y = 2[/latex], or at the coordinate [latex](-2, 2).[/latex]

Find the solution to the following system of equations: [latex]2x+3y=-4[/latex] and [latex]3x-4y=11.[/latex]

First, line up the equations and choose the variable that shall be eliminated:

[latex]\left\{ \begin{array}{rrrrr} 2x&+&3y&=&-4 \\ 3x&-&4y&=&11 \end{array}\right.[/latex]

It looks the simplest to eliminate the [latex]x[/latex] variable. This means the top equation needs to be multiplied by 3 and the bottom equation multiplied by −2. Then, the two equations are added together, and each side is divided by 17:

[latex]\begin{array}{rrrrrrl} &(2x&+&3y&=&-4)&(3) \\ &(3x&-&4y&=&11)&(-2) \\ \\ &6x&+&9y&=&-12& \\ +&-6x&+&8y&=&-22& \\ \hline &&&\dfrac{17y}{17}&=&\dfrac{-34}{17}& \\ \\ &&&y&=&-2& \end{array}[/latex]

It is now known that these equations intersect at the value where [latex]y = -2.[/latex] Now choose one of the two original equations (choose [latex]2x + 3y = -4[/latex]) and substitute [latex]y = -2[/latex] to find x:

[latex]\begin{array}{rrcrr} 2x&+&3(-2)&=&-4 \\ 2x&-&6&=&-4 \\ &+&6&&+6 \\ \hline &&\dfrac{2x}{2}&=&\dfrac{2}{2} \\ \\ &&x&=&1 \end{array}[/latex]

The intersection point of these two linear equations is [latex]x = 1[/latex] and [latex]y = -2[/latex], or at the coordinate [latex](1, -2).[/latex]

Find the solution to the following system of equations: [latex]2x-5y=3[/latex] and [latex]-6x+15y=-9.[/latex]

First, line up the equations to choose the variable that shall be eliminated:

[latex]\left\{ \begin{array}{rrrrr} 2x&-&5y&=&3 \\ -6x&+&15y&=&-9 \end{array}\right.[/latex]

To eliminate the [latex]x[/latex] variable, the top equation needs to be multiplied by 3:

[latex]\begin{array}{rrrrrrr} &(2x&-&5y&=&3)&(3) \\ \\ &6x&-&15y&=&9& \\ +&-6x&+&15y&=&-9& \\ \hline &&&0&=&0& \\ \end{array}[/latex]

Everything cancels out because the two equations are identical. Therefore, there are infinite solutions.

Find the solution to the following system of equations: [latex]4x-6y=8[/latex] and [latex]4x-6y=-4.[/latex]

Once these two equations are aligned, it is easy to see they are identical except for their intercepts. They are parallel lines. To cancel the variables out, one of the two equations must be multiplied by −1:

[latex]\begin{array}{rrrrrrr} &4x&-&6y&=&8& \\ &4x&-&6y&=&-4&(-1) \\ \\ &4x&-&6y&=&8& \\ +&-4x&+&6y&=&4& \\ \hline &&&0&=&12& \\ \end{array}[/latex]

The result is all the variables cancelling out to 0 and falsely equalling some number. There is no solution, since these equations will never intercept each other.