Solving Systems of Inequalities

Solve the following systems of inequalities.

$\left\{\begin{array}{l}y\le x-1\\ y<–2x+1\end{array}\right.$

Solution

Since we already have the y variable isolated in both inequalities, we will go ahead and graph that immediately. Let us find the points we would have to graph them with. We will use the intercept method here. What will be the value of x when y = 0? What will be the value of y, when x = 0? We can replace the inequality sign with an equation sign so it gets easier to solve for now.

When $x=0$,

$y=x-1$

$y=0-1$

$y=-1$

$(0,-1)$

When $y=0$,

$y=x-1$

$0=x-1$

$x=1$

$(1,0)$

We now have coordinates for our first line. However, because the sign there is ≤, the line of the graph will be solid. We can also determine which side of the line will have to be shaded mathematically by substituting (0, 0) into the equation to see if it is true.

$y\le x-1$

$0\le 0-1$

$0\le -1$

This means that the point (0, 0) is not less or equal to -1, therefore, we will shade the opposite side of the line where (0, 0) does not exist.

We will graph the second inequality also by finding two points using the intercept method. What will be the value of x when y = 0? What will be the value of y, when x = 0? We can replace the inequality sign with an equation sign so it gets easier to solve for now.

$y=-2x+1$

When $x=0$,

$y=-2\left(0\right)+1$

$y=1$

$(0,1)$

When $y=0$,

$0=-2\left(x\right)+1$

$-2x=1$

$x=-0.5$

$(-0.5,0)$

We now have coordinates for our second line. However, because the sign there is <, the line of the graph will be dotted. We will also determine which side of the line will have to be shaded mathematically by substituting (0, 0) into the equation to see if it is true.

$y<-2x+1$

$0<-2\left(0\right)+1$

$0<1$

This is actually true, therefore we will shade the part of the line that has the point (0, 0).

The solution of the system is the intersection of the two shaded regions.

Solve the following system of inequalities.

$\left\{\begin{array}{l}6x-2y\ge 12\\ 3x+4y>12\end{array}\right.$

Solution

We will graph the first inequality first. We will find the points by using the intercept method.

$6x-2y=12$

When $x=0$,

$6\left(0\right)-2y=12$

$y=-6$

$(0,-6)$

When $y=0$,

$6x-2\left(0\right)=12$

$x=2$

$(2,0)$

Since we have enough points to construct the line, we will plot our first inequality.

We will graph the second inequality also by finding two points using the intercept method.

$3x+4y=12$

When $x=0$,

$3\left(0\right)+4y=12$

$(0,3)$

When $y=0$,

$3x+4\left(0\right)=12$

$x=4$

$(4,0)$

The solution of the system is the intersection of the two shaded regions.

Solve the following system of inequalities.

$\left\{\begin{array}{l}-4x+6y>6\\ 2x-3y>3\end{array}\right.$

Solution

Let us first graph the first inequality by using the intercept method.

When $x=0$,

$-4\left(0\right)+6y=6$

$y=1$

$(0,1)$

When $y=0$,

$-4x+6\left(0\right)=6$

$x=-1.5$

$(-1.5,0)$

Since we have enough points to construct the line, we will plot our first inequality.

We will graph the second inequality also by finding two points using the intercept method.

$2x-3y=3$

When $x=0$,

$2\left(0\right)-3y=3$

$y=-1$

$(0,-1)$

When $y=0$,

$2x-3\left(0\right)=3$

$x=1.5$

$(1.5,0)$

We notice here that both lines are parallel, hence, there is no region that intersects. These are called systems with no solutions.

Solve the inequality below and represent it on a number line.

$\left\{\begin{array}{l}2x+3\ge 1\\ -x+2\ge -1\end{array}\right.$

Solution

As mentioned earlier, we will solve each inequality separately. So we will take the first inequality here.

We will now solve this algebraically, in an attempt to isolate the x variable. By that, we will subtract 3 from each side of the inequality.

$2x+3-3\ge 1-3$

$2x\ge -2$

Divide both sides of the inequality by 2 to isolate the x.

$\frac{2x}{2}\ge \frac{-2}{2}$

$x\ge -1$

The interval notation will be written as $[-1,\infty )$

We now have a solution for the first inequality. Let's do the same process for the second.

$-x+2\ge -1$

We will also want to isolate the x variable in this inequality too. We will subtract 2 from each side of the inequality.

$-x+2-2\ge -1-2$

$-x\ge -3$

We can now simply multiply each side of the inequality by –1. However, a rule on dealing with inequalities says that the sign changes to be the opposite once both sides are multiplied by a negative number. Hence, ≥ will become ≤.

$-1(-x)\ge -1(-3)$

$x\le 3$

Notice that the sign changes above?

The interval notation will be written as $(\infty ,3]$

The intersection of these solution sets is the set;

$[-1,3]$

Solve the inequality below and write the interval notation of it.

$\left\{\begin{array}{l}2x+3<1\\ -x+6<3\end{array}\right.$

Solution

We will solve both inequalities separately. We will do the first one first.

$2x+3<1$

We will attempt to isolate the y by first subtracting 3 from each side of the inequality.

$$\begin{gathered} 2x+3-3<1-3 \\ 2x<-2 \end{gathered}$$

We will divide each side of the inequality by 2.

$$\begin{gathered} \frac{2x}{2}<\hspace{0.17em}\frac{-2}{2} \\ x<-1 \end{gathered}$$

The solution set in interval notation is $(\infty ,-1)$.

We will now solve the second inequality.

$-x+6<3$

We will isolate x by subtracting 6 from each side of the equation

$$\begin{gathered} -x+6-6<3-6 \\ -x<-3 \\ -1(-x)<-1(-3) \end{gathered}$$

We will multiply each side of the inequality by –1. The sign changes to be the opposite once both sides are multiplied by a negative number. Hence, < will become >.

$x>3$

The solution set in interval notation is $(3,\infty )$.