Elementary Algebra

We are looking for the point where the two \(y\)-values are equal. Therefore, we set the two expressions for \(y\) equal, which gives us an equation in \(x\) to solve:

We find that \(x=50\text{,}\) the same answer we got using graphing. Substituting \(x=50\) into either equation gives \(y=1300\text{.}\) The solution is \((50, 1300)\text{.}\)

We first solve one of the equations for \(y\) in terms of \(x\text{.}\) It is easier to solve the first equation for \(x\text{,}\) by subtracting \(3x\) from both sides to get:

We now substitute the expression \(1-3x\) for \(y\) in the other equation:

This is the \(x\)-coordinate of the solution point. To find the \(y\)-coordinate, we substitute \(x=\alert{-3}\) into either of the original equations. It is easiest to use the equation we solved for \(y\text{,}\) namely \(y=1-3x\text{.}\)

The solution to the system is \(x=-3,~ y=10\text{,}\) or the point \((-3,10)\text{.}\) You can check that these values satisfy both of the original equations in the system.

First, we rewrite each equation in the form \(Ax+By=C\text{.}\)

We add the equations together by adding the left side of the first equation to the left side of the second equation, and then adding the two right sides together, as follows:

Note that the \(y\)-terms canceled, or were eliminated. We are left with an equation in \(x\) that is easy to solve:

We are not finished yet, because we must still find the value of \(y\text{.}\) We can substitute our value for \(x\) into either of the original equations, and solve for \(y\text{.}\) We'll use the second equation, \(3x+2y=19\text{:}\)

Thus, the solution is the point \((5,2)\text{.}\)

We can choose to eliminate either the \(x\)-terms or the \(y\)-terms in a system. For this example, it will be faster to eliminate the \(y\)-terms. If we multiply each term of the second equation by \(-3\text{,}\) then the coefficients of \(y\) will be opposites:

Note carefully that we must multiply each term by \(-3\text{,}\) not just the \(y\)-term. We can then replace the second equation by its new version to obtain a new system, and we add the equations together:

The \(y\)-terms were eliminated, and we solve the resulting equation for \(x\) to get \(x=-2\text{.}\) Finally, we substitute \(x=-2\) into either of the equations to find \(y=5\text{.}\) The solution is \((-2,5)\text{.}\)

This time we choose to eliminate the \(x\)-terms. We must arrange things so that the coefficients of the \(x\)-terms are opposites, so we look for the smallest integer that both 2 and 5 divide into evenly. (This number is called the lowest common multiple, or LCM, of 2 and 5.) The LCM of 2 and 5 is 10.

We want one of the coefficients of \(x\) to be 10, and the other to be \(-10\text{.}\) To achieve this, we multiply the first equation by 2 and the second equation by \(-5\text{.}\)

Adding these new equations eliminates the \(x\)-term and yields an equation in \(y\text{.}\)

We solve for \(y\) to find \(y=-1\text{.}\) Finally, we substitute \(y=\alert{-1}\) into the first equation and solve for \(x\text{.}\)

The solution to the system is \((4,-1)\text{.}\)

1. To eliminate the \(y\)-terms, we multiply the first equation by 2 and add:

   \begin{align*} 6x-2y \amp = 4 \\ \underline{-6x+2y}\amp\underline{{}= 3\vphantom{y}}\\ 0x+0y \amp = 7 \end{align*}

   

   Both variables are eliminated, and we are left with the false statement \(0=7\text{.}\) There are no values of \(x\) or \(y\) that will make this equation true, so the system has no solutions. The graph shows that the system is inconsistent.

2. To eliminate the \(x\)-terms, we multiply the first equation by \(-2\) and add:

   \begin{align*} -2x+4y \amp = -6\\ \underline{2x-4y}\amp \underline{{}= \hphantom{-}6\vphantom{y}}\\ 0x+0y \amp = 0 \end{align*}

   

   We are left with the true but unhelpful equation \(0=0\text{.}\) The two equations are in fact equivalent (one is a constant multiple of the other), so the system is dependent. The graph of both equations is shown in the figure.