RWM102 Study Guide

6a. Determine the number of solutions of a given system of linear equations

• How do you verify if a point is a solution to a system of equations?

A solution to a system of equations is just like the solution to a single linear equation, except that the point must satisfy both equations in order to be considered the solution to the system of equations.

For example, the system of equations:

$x+2y=13$

$3x-y=-11$

Let's check if the point (-1,7) is a solution. To check, first we will substitute the point into the first equation.

$-1+2(7)=-1+14=13$

So far, the point works, but we must make sure it works in the other equation as well:

$3(-1)-7=-3-7=-10$

Since this does not satisfy both equations, (-1,7) is not a solution to this system.

To review, see Checking Solutions for Systems of Linear Equations

6b. Classify systems of linear equations according to the number of solutions

• How do you know the number of solutions of a system of linear equations?

Systems of linear equations can have 0, 1, or infinite solutions. Most commonly, two lines intersect at only one point, meaning the system has 1 solution. If the two lines are parallel, then they never intersect, and therefore the system has no solution. Finally, if the system has two equations that are actually representative of the same line, then all the points on each line are also a solution to the other equation, meaning there are infinitely many solutions.

To review, see Using Graphs to Solve Linear Equations.

6c. Solve systems of linear equations using graphing, substitution, or elimination

• How do you solve a system of linear equations with graphing?
• How do you solve a system of linear equations with substitution?
• How do you solve a system of linear equations with elimination?

Systems of linear equations can be solved through 3 methods, each with advantages and disadvantages.

First, systems of linear equations can be solved by graphing. To use graphing, you only need to graph each line on the same coordinate plane, and then find the point where the lines cross. That point is the solution to the system.

For example, consider the following system of equations:

$y=3x-1$

$y=-x+3$

We can graph both lines and look for the point where they intersect.

Since the lines intersect at (1,2), that is the solution to the system. The graphing method works well when the solution is a lattice point, with whole number values, but is not as effective if the answers are fractions or decimals.

Another method is substitution. Substitution is an algebraic method, rather than the geometric method of graphing. In substitution, we solve one equation for either $x$ or $y$, and then substitute that value into the other equation to find the value of one variable. Once we have found that value, we can substitute it to find the value of the other variable.

For example, let us once again consider our example:

$y=3x-1$

$y=-x+3$

Since $y=-x+3$ in the second equation, we can replace the $y$ in the first equation with that value:

$-x+3=3x-1$

Now we can solve for $x$. $4=4x$, therefore $x=1$. We can now use that value to find the value of $y$: $y=3(1)-1=2$. Therefore the solution is (1,2)

Finally, we can solve a system of equations by elimination. This method is best for systems where one variable can't be isolated that easily. In this method, we multiply one or both equations so that when we add them together, one of the variables cancels. This will allow us to solve for one variable, and then as we did with substitution, we can use that value to find the other remaining value.

For example consider the following system of equations:

$3x+2y=7$

$-x-4y=6$

First, we will multiply the top equation by 2, so that when we add the equations, the $y$ terms will cancel:

$6x+4y=14$

$-x-4y=6$

Now we add the two equations together and solve for $x$:

$5x=20$, $x=4$

Now that we know $x=4$, we can substitute into one of the original equations to find $y$:

$3(4)+2y=7$

Now we can solve for $y$:

$2y=-5$, $y=-52$

Therefore the solution to this system of linear equations is (4, -52)

6d. Locate on a coordinate plane all solutions of a given system of inequalities

• How do you graph the solutions to a system of linear inequalities?

When solving a system of inequalities, graph the solution to each inequality, and shade the side with the solutions. When you have done both, look for the area where the shading overlaps. That is the area with the solutions that work for both inequalities, and are therefore the solutions to the system of inequalities.

For example:

To review, see Using Graphs to Solve Linear Equations

6e. Create systems of equations and use them to solve world problems

• How do you use systems of equations to solve real-world problems?

As we have seen, systems of equations are helpful in solving real-world problems. When given a real-world problem, we can create a system of equations to find the solution. For example, consider the following problem:

Juan is considering two cell phone plans. The first company charges $120 for the phone and$30 per month for the calling plan that Juan wants. The second company charges $40 for the same phone but charges$45 per month for the calling plan that Juan wants. After how many months would the total cost of the two plans be the same?

First, we need to create two linear equations to represent the problem:

First company: $120+30x=y$

Second company: $40+45x=y$

Since 120 and 40 are the fixed costs, they are the constants, and the monthly cost is the coefficient of $x$, since each month you have to pay that amount.

Now we can solve this system of equations in any of the ways we have already learned, such as elimination, substitution, or graphing.

In this case, we can use substitution to get $x$:

$120+30x=40+45x$

We can solve this to find $80=15x$, $x=\dfrac{80}{15}=5\dfrac{1}{3}$ months.

6f. Use systems of inequalities to model word problems and interpret their solutions in the context of the problem

• How can you use systems of inequalities to solve word problems?

Sometimes, a system of equations isn't appropriate for our problem. Instead, a system of inequalities should be used. For example, consider the following problem:

Jake does not want to spend more than $50 on bags of fertilizer and peat moss for his garden. Fertilizer costs$2 a bag and peat moss costs \$5 a bag. Jake's van can hold at most 20 bags.

First, we must create our inequalities. We will let Fertilizer $= x$, and Peat Moss $= y$:

$x+y \leq 20$

$2x+5y \geq 50$

In addition to these two inequalities that we can create from the problem, remember that $x \geq 0$ and $y \geq 0$, since Jake obviously can't have a negative number of bags of something.

We can now graph the solution to this system and then interpret the answers:

As you can see in the solution above, the area with the diagonal lines is the solution to our system of equations. Every point in that area is a solution. For example, (5,5) is a solution, meaning Jake could buy 5 bags of fertilizer and 5 bags of peat moss.

Unit 6 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• system of equations
• intersect
• parallel
• graphing
• lattice point
• substitution
• elimination
• system of inequalities