Algebra Toolkit

We begin just as we do to solve an equation. The only difference is that we must reverse the direction of the inequality if we multiply or divide by a negative number.

In interval notation, the solution set is \((-6,\infty)\text{.}\)

1. This equation is in the form \(Ax+By=C\text{,}\) so the intercept method of graphing is convenient. The intercepts are \((150,0)\) and \((0,-75)\text{.}\) The graph is shown below.

   

2. This equation is in the form \(y=mx+b\text{,}\) so the slope-intercept method of graphing is convenient. The \(y\)-intercept is \((0,400)\text{,}\) and the slope is \(-25\text{.}\) The graph is shown below.

   

We start by solving the second equation for \(x\) to get \(x=2y-2\text{.}\) Then we substitute this expression for \(x\) into the first equation, which gives us

We solve this equation for \(y\) to find \(y=1\text{.}\) Finally, we substitute \(y=1\) into our first step to find

The solution is \(x=0,~y=1\text{,}\) or \((0,1)\text{.}\)

We multiply the first equation by 3 and the second equation by \(-2\) in order to eliminate \(x\text{.}\)

Adding these two equations gives us \(-y=1\text{,}\) or \(y=-1\text{.}\) Finally, we substitute \(y=-1\) into either equation (we choose the first equation), and solve for \(x\text{.}\)

We find\(x=1\text{,}\) so the solution is \(x=1,~y=-1\text{,}\) or \((1,-1)\text{.}\)