MA001 Study Guide

9a. solve systems of linear equations in two and three variables using graphs, substitution, and addition

• Compare solving systems of linear equations in two variables via graphs, substitution, or addition.
• How might graphs, substitution, and addition be used to solve systems of linear equations in three variables?

There are often contexts that require the use of more than one equation to be solved at the same time. A system of linear equations consists of two or more linear equations with two or more variables that must be solved at the same time. The solution is the ordered pair or ordered triple whose values for the variables satisfy each equation at the same time. For a system to have the potential for a unique solution, there should be as many linear equations as there are variables.

Consider the following system of three linear equations in three variables.

$\begin{aligned} &2 x+5 y-7 z=-39 \\ &4 x+y+z=3 \\ &9 x-2 y-2 z=11 \end{aligned}$

The ordered triple $(1, \, -4, \, 3)$ is a solution to the system. When the coordinates are substituted into each of the three equations, each equation is a true statement.

There are three techniques that are often used to solve a linear system. Consider the system below of two linear equations in two variables.

$\begin{aligned} &5 x-4 y=-3 \\ &6 x+y=8 \end{aligned}$

The first technique is to graph both lines and then find the coordinates of the point of intersection. The coordinate grid below shows the graph of both lines. The two lines intersect at the point $(1, \, 2)$, which is the solution to the system.

Sometimes it is not easy to graph the lines or the coordinates of the intersection are not integers. In this situation, it helps to have algebraic techniques. One algebraic technique is the substitution method. Solve for one of the variables and then substitute for that variable in the other equation and solve. For the system above, the second equation can be easily solved for $y$ to obtain $y=8-6 x$. Now substitute this value for $y$ into the first equation.

$5 x-4(8-6 x)=-3 \text { leads to } 5 x-32+24 x=-3$

Then $29 x=29$, so $x=1$. Substitute this value for $x$ into $y=8-6 x$ to get $y=2$. So the solution is $(1, \, 2)$ as found on the graph.

If solving for one of the variables leads to fractional coefficients, it may be more efficient to use a second algebraic technique, specifically the addition method. The goal here is to modify one or more of the equations so that the coefficients of one of the variables are opposites of each other. Then, when the equations are added, this variable is eliminated and a solution to the other variable can be found. For the system earlier, it might help to adjust the second equation so that the coefficient for the variable $y$ is $-4$, the opposite of the coefficient of the variable $y$ for the first equation. Multiplying the second equation by 4 yields $24 x+4 y=32$. When this equation is added to the first equation, the result becomes $29 x=29$, so $x=1$. Substituting this value into either equation yields $y=2$, for the solution $(1,2)$ found by both of the other two methods.

For the addition method, it is possible to adjust the equations of the previous system so that the coefficients of the variable $x$ are opposites; given the coefficients of 5 and 6 from the two equations, the least common multiple is 30. So, multiply the first equation by 6 to get $30 x-24 y=-18$ and multiply the second equation by $-5$ to get $-30 x-5 y=-40$. Adding these two equations yields $-29 y=-58$, so $y=2$. Substituting into either equation then gives $x=1$. Although it does not matter which variable is eliminated when adding, sometimes eliminating one variable is more efficient than eliminating the other.

The same three techniques can be used to solve a system of three linear equations in three variables. However, these techniques are more complicated when three variables are involved. The solution to a system of three linear equations in three variables is an ordered triple $(x, y, z)$. So, trying to graph the solution means trying to graph in 3-space, which is hard to accomplish on paper. The substitution method is not very efficient. So, generally the addition method is used to solve a system of three linear equations in three variables. However, this method will need to be used more than once. Take two of the equations and use the addition method to eliminate one of the variables. Then take a different two equations and eliminate the same variable. The result is a system of two equations in two variables, and so the addition method is applied again.

Review material in Analyzing the Solution to a System in Two Variables and Solve Systems with Three Variables.

9b. categorize solutions of systems of linear equations

• Compare the solution in a consistent system and an inconsistent system. 
• Contrast the solution in an independent system and a dependent system. 

A consistent system has at least one solution; an inconsistent system has no solution. There are two types of consistent systems; an independent system has a single solution whereas a dependent system has an infinite number of solutions. For a system of two linear equations in two variables, these different types of systems can be described graphically. An independent system has a single solution so it consists of two lines with different slopes that intersect in a single point. A dependent system consists of two lines with the same slope and the same y-intercept; that is, the two lines are coincident so that every point on one line satisfies the second line. This means that the solution to a dependent system of two linear equations is actually a line. An inconsistent system consists of two lines with the same slope but different y-intercepts; the two lines are parallel and never intersect. When solving algebraically, if the process leads to a statement that is always true, then the system is a dependent system; if the process leads to a statement that is false, then the system is an inconsistent system. 

Similar analysis describes the nature of the solutions to three linear equations in three variables. An independent system consists of three planes that intersect in a single point with coordinates $(x, \, y, \, z)$. A dependent system of three linear equations consists of three coincident planes, three planes that intersect in a line, or two coincident planes that intersect the third plane in a line. A dependent system can be recognized in an algebraic process when a true statement is obtained. An inconsistent system has no solution and consists of three parallel planes, two parallel lines that intersect with a third plane, or three planes in which each plane intersects the other two planes but the intersections are not at the same point. An inconsistent system can be recognized in an algebraic process when a false statement is obtained.

Review material in Analyzing the Solution to a System in Two Variables and Solve Systems with Three Variables.

9c. solve systems of nonlinear equations using substitution and elimination

• Be able to use substitution or elimination to solve a nonlinear system of equations in two variables.

At times, a system of two equations in two variables may consist of an equation that is not a linear equation, that is, one of the variables has a power greater than 1 or the two variables are multiplied. Such a system is a system of nonlinear equations. However, the techniques of substitution and addition (elimination) used with linear systems can still be applied. As with linear systems, there are several possibilities for the solution.

One of the simplest nonlinear systems consists of a quadratic equation and a linear equation, that is, an equation whose graph is a parabola and an equation whose graph is a line. In this case, three possibilities exist: the line might intersect the parabola in two points; the line might be tangent to the parabola, intersecting it in just one point; or the line might not intersect the parabola at all so that there is no solution.

Consider the system below and its graph.

$\begin{aligned} &y=-2 x+10 \\ &y=(x-2)^{2}+3^{\square} \end{aligned}$

Substituting the value for $y$ from the line into the value of $y$ for the parabola leads to the following.

$-2 x+10=(x-2)^{2}+3$ or $x^{2}-2 x-3=0$ leading to $x=3$ or $x=-1$

So the solutions are the ordered pairs $(3,4)$ and $(-1,12)$. Both ordered pairs are visible on the graph and satisfy both equations.

A second common nonlinear system consists of a circle and a line. Again, there are three possibilities: two solutions if the line intersects the circle in two points; one solution if the line is tangent to the circle; and no solutions if the line does not intersect the circle.

When a line is one of the equations in the nonlinear system, substitution is typically the most efficient method to use. However, if the nonlinear system consists of two parabolas or two equations with both variables to a power greater than 1 (e.g., the equation of a circle), then it is likely more efficient to use addition to eliminate one of the variables, just as is done with a linear system.

Review material in Algebraic Methods for Solving Systems of Non-Linear Equations.

9d. graph single and systems of nonlinear inequalities

• Compare graphing an inequality to graphing an equation.
• Explain how to graph the solution to a system of inequalities.

The graph of an equation consists of those ordered pairs that make the equation true. In contrast, the graph of an inequality consists of a region. In fact, the graph of an equation $y=f(x)$ divides the coordinate grid into three regions: those ordered pairs that make the equation true, that is, ordered pairs for which $y=f(x)$; those ordered pairs that when substituted into $f(x)$ lead to $y < f(x)$; and those ordered pairs that when substituted into $f(x)$ lead to $y > f(x)$. If $y \leq f(x)$ or $y \geq f(x)$ then ordered pairs on the equation are also part of the solution; in this case, the equation is graphed with a solid curve. If the inequality is a strict inequality (that is, $<$ or $>$), then the equation is graphed with a dashed curve.

To graph an inequality, first graph the equation associated with the inequality, using either a solid or dashed curve as appropriate. Then choose a point not on the equation and substitute into the inequality. If the inequality is true, then the region containing that ordered pair is the solution; if the inequality is not true, then the region on the other side of the equation is the solution.

For instance, the graph shown is the graph of $y < x^{2}$. The ordered pair $(0,1)$ leads to $1 < 0$ which is false; so, the region containing $(0,1)$ is not part of the solution. In contrast, the ordered pair $(2,0)$ leads to $0 < 4$ which is true; so, the solution is the region containing $(2,0)$, denoted by shading the region (the purple region). The parabola itself is dotted to indicate that ordered pairs on the parabola are not part of the solution.

A system of inequalities is a system containing inequalities rather than equations. To solve such a system, graph the solution to each inequality separately, and then determine where the solutions overlap. The solution to the system is represented by the overlap.

Consider the system of inequalities graphed below.

$x^{2}+y^{2} < 4^{\square}$and $y < x-2$

The solution to the system is the dark green region that is both inside the circle and below the line.

Review material in Algebraic Methods for Solving Systems of Non-Linear Equations.

9e. construct an augmented matrix from a system of linear equations

• Identify an augmented matrix for a system of linear equations.
• Given an augmented matrix, write a system of linear equations.

An augmented matrix for a system is a matrix that contains the coefficients of the variables and the constants, with a vertical line used to separate the coefficients from the constants. For a system of two equations in two variables, the matrix is a $2 \times 3$ matrix, meaning there are 2 rows and 3 columns. For a system of three equations in three variables, the matrix is a $3 \times 4$ matrix, meaning there are 3 rows and 4 columns.

Consider systems $A$ and $B$ below.

A. $\begin{array}{ll} 5 x+7 y=-11 & \text { B. } 4 x-y+2 z=5 \\ 3 x+2 y=8 & 7 x+4 y-2 z=3 \\ & 7 x \end{array}$

The augmented matrices for these two systems are shown below.

A. $\left[\begin{array}{cc|c}5 & 7 & -11 \\ 1 & -2 & 8\end{array}\right]$                      B. $\left[\begin{array}{ccc|c}4 & -1 & 2 & 5 \\ 3 & 1 & 1 & 2 \\ 7 & 4 & -2 & 3\end{array}\right]$

It is also possible to write a system when given the augmented matrix. For instance, given the matrix below.

$\left[\begin{array}{rr|c} 9 & 4 & -6 \\ 5 & -1 & -13 \end{array}\right]$

The associated system consists of the two equations $9 x+4 y=-6$ and $5 x-y=-13$.

Once an augmented matrix is obtained, it can be used to solve a system by working only with the coefficients. Using the same procedures as used in the addition method, work on the rows of the matrix to get the matrix in a form which has $1 \mathrm{~s}$ on the diagonal and 0s below the diagonal. Then the values of the variables can be identified. For instance, for the augmented matrix above, apply the following steps.

• Multiply the first row by $\frac{1}{9}$ to obtain the matrix $\quad\left[\begin{array}{cc|c}1 & \frac{4}{9} & \frac{-6}{9} \\ 5 & -1 & -13\end{array}\right]$
• Now replace the second row with the result of multiplying the first row by $-5$ and adding it to the second row to obtain the matrix $\left[\begin{array}{ccc} 1 & \frac{4}{9} & -6 \\ 0 & \frac{-29}{9} & -87 \\ 0 & \frac{-87}{9} \end{array}\right]$
• Solve for the variables. $-\frac{29}{9} y=-\frac{87}{9}$, so $y=3$. Substituting this value of $y$ into the equation of the first row leads to $x+\frac{4}{9} \cdot 3=-\frac{6}{9}$, so $x=-2$. The solution to the system is the ordered pair $(-2, \, 3)$.

Review the material in Algebraic Methods for Solving Systems of Non-Linear Equations.

Unit 9 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• addition method for solving systems
• augmented matrix
• consistent system
• dependent system
• inconsistent system
• independent system
• solution to a system
• substitution method for solving systems
• system of inequalities
• system of linear equations
• system of nonlinear equations