Solving Systems of Equations

In an extremely abstract sense, a system of equations is collection of equations with potentially related variables. There are numerous interpretations and uses for systems of equations. Another frequent interpretation of a system of equations is that each equation describes a geometric object and a solution (or lack thereof) describes the intersection(s or lack thereof) of these geometric objects.

The most elementary objects that we might consider are lines. $2x + y = 1$ and $3x + y = 1$ represent a pair of lines. By considering the x and y variables to exist on the same cartesian plan, we then imagine that either they intersect or they do not. If we consider the figure to the right, we notice that they do, in fact intersect. It turns out it is easy to generalize a line. Lines may live in 3 or 4 or n dimensions. We might also consider objects like planes that behave a bit like lines if we consider them subsets of higher dimensional spaces. $2x + 3y + 2z = 5$ describes a plane in 3 dimensions. There are several algebraic techniques to solve for solutions to sets of linear equations.

Note that in most problems that present systems of linear equations, these equations are often given in “standard form”: $Ax+By=C$. Depending on the method you choose to find the solution, one might prefer this form or need to transform the linear equation into “slope-intercept” form: $y=mx+b$.

There are several elegant and simple methods for solving linear systems of equations. This becomes the gateway to a modern field of mathematics that you will soon study known as linear algebra. Linear algebra a sprawling and enormously successful field of mathematics.

There are some special cases to be aware of when solving systems of equations. A system of equations will have infinite solutions when the lines are exactly the same; recall that this happens when two lines have both the same slope and the same y-intercept. Consider the following example:

A system of equations will have no solution when they never intersect. Recall that two lines never intersect in the x-y plane if they are parallel, and that two lines will be parallel if they have the same slope but different y-intercepts. Consider the following example:

Let’s consider a more nuanced example, where we might seek to make a system have no or infinitely many solutions given some unknown quality.

Systems of linear inequalities add only very little to the concepts we have covered herein. We are essentially combining linear inequalities with the concepts from systems of linear equations. In Systems of Linear Inequalities, we are considered not only a region of solutions, but the overlapping of two regions of solutions; this overlapping region is the set of solutions to the system of linear inequalities. 

For example, consider the graph below

First, let’s find the function that describes each line, not considering inequalities.

• The blue line has a y-intercept of $(0, 3)$, so our $b$ value for our slope-intercept form will be 3. Picking some convenient points – $(0, 3)$ and $(-3, 1)$ – we can find the slope

$$m=\frac{\Delta{y}}{\Delta{x}}=\frac{3-1}{0-{-3}}=\frac{2}{3}$$

Thus, the equation of the blue line is $y_{\text{blue}}=\frac{2}{3}x+3$

• The red line has a y-intercept of $(0, -1)$, so our $b$ value for our slope-intercept form will be -1. Picking some convenient points – $(0, -1)$ and $(5, -5)$ – we can find the slope

$$m=\frac{\Delta{y}}{\Delta{x}}=\frac{-5-{-1}}{5-0}=\frac{-4}{5}$$

Thus, the equation of the red line is $y_{\text{red}}=\frac{-4}{5}x-1$

Now let us consider the implied inequalities for each function given the graph. 

The blue line is solid, meaning that the set of solutions does exist on the line itself; additionally, there is a blue field below the blue line, so we should assume that solutions also exist for values “less than” the function. Thus, we can say for the blue function that

$$y_{\text{blue}}\le\frac{2}{3}x+3$$

The red line is dashed, meaning that the set of solutions does NOT exist on the line itself; additionally, there is a red field above the red line, so we should assume that solutions do exist for values “greater than” the function. Thus, we can say for the red function that

$$y_{\text{red}}\gt\frac{-4}{5}x-1$$

When we graph these functions and fields as given in the diagram above, we find that the two fields overlap and form a third region. This overlapping region is the set of solutions. However, we need to define this region’s domain (or possible x-values) and range (possible y-values).

Looking at the graph, we see that the point of intersection of the two functions is the left most point of the solution field. Thus, we need to find the “solution” to the system of linear equations to get this point. Let’s do this! Since we have both equations in slope-intercept form, they both equal $y$, and so we can use the “Equivalency Method” to set them equal to each other and solve for $x$.

$$\frac{2}{3}x+3=\frac{-4}{5}x-1$$

$$\frac{2}{3}x-\frac{-4}{5}x+3=\frac{-4}{5}x-\frac{-4}{5}x-1$$

$$\frac{10}{15}x-\frac{-12}{15}x+3=-1$$

$$\frac{10}{15}x+\frac{12}{15}x+3=-1$$

$$\frac{22}{15}x+3=-1$$

$$\frac{22}{15}x+3-3=-1-3$$

$$\frac{22}{15}x=-4$$

$$22x=-15$$

$$x=\frac{-15}{22}$$

Now solve for $y$ by substituting this value of $x$ back into one of the functions.

$$y=\frac{2}{3}x+3\rightarrow{y=\frac{2}{3}(\frac{-15}{22})+3}$$

$$y=\frac{2(-15)}{3(22)}+\frac{3(22)}{22}$$

$$y=\frac{-30}{66}+\frac{198}{66}$$

$$y=\frac{-168}{66}=\frac{-28}{11}$$

Thus, the point of intersection for the two linear inequalities is $(\frac{-15}{22}, \frac{-28}{11})$ . Consequently, the domain of our solutions must be greater than this point, but also equal to this point, since the blue function allows solutions on its line. Thus $x\ge\frac{-15}{22}$ is the domain of our solutions.

The range of our solutions must lie between the blue line and the red line as $x$ goes to positive $\infty$. Furthermore, the range can go up to and include the blue line, but cannot include the red line as it goes down; thus, it is less than or equal to the blue function but is only greater than (not equal to) the red function. We can write this as $\frac{-4}{5}x-1\lt{y}\le\frac{2}{3}x+3$

Thus the field of all solutions to this system of two linear inequalities is:

$$x\ge\frac{-15}{22}$$

$$\frac{-4}{5}x-1\lt{y}\le\frac{2}{3}x+3$$

We can apply what we have learned in systems of linear equations and inequalities to optimize solutions to linear systems (often called “linear optimization”). Linear Optimization is the process of finding a maximum or minimum value for some function under certain conditions. No new skills are required here; we are merely applying what we have learned. Consider the below example:

Example: Find the maximum value of $y$ given $–3x + 2y \le4$ and $x + y \le 1$ subject to the condition that $x \ge 0$.

First, let’s put these two inequalities in slope-intercept form so that we can graph them:

$$–3x + 2y \le4$$

$$ 2y \le3x+4$$

$$ y \le\frac{3}{2}x+2$$

$$x + y \le 1$$

$$y \le {-x}+1$$

Now we know that $ y \le \frac{3}{2}x+2$ (the red line and red region) and $y \le {-x}+1$ (blue line and blue region) with the additional condition that $x \ge 0$. In real-world applications, we are often only considering when $x \ge 0$ (y-axis and yellow region) due to considerations of real-world quantities that do not often come in the negative, such as time and population. When we graph these functions and their appropriate inequality regions, we get this graph.

All the lines must be solid since values equal to the functions are also valid. In this example, we have three overlapping solution regions, since $x \ge 0$ also shades the entire region to the right of the y-axis in yellow.

The darkest shaded region (the wedge in the lower right of the graph) satisfies all the constraints on the problem – it is where the red, blue, and yellow regions all overlap. We then want to find the maximum value of y, which is clearly 1. We can also find this value by substituting $x = 0$ into $ y \le {-x} + 1$ and finding the maximum value of y, which likewise is clearly 1.

Sometimes you will be asked to create and solve a system of equations that models a real-world situation. The first thing that you should always do when approaching these problems is read the question carefully and define your variables. You will often notice recurring themes in such questions:

1. Usually you will find that you have two varieties of some things, which we could designate as $x$ and $y$ (e.g. chairs and tables as two types of furniture, adult tickets and children tickets as two types of tickets, apples and oranges as two types of fruit, etc.). 
2. You will also find that these two kinds of things have some “weight” or “value,” which we will append to the things: that is, $x$ as some “weight” $a$, and $y$ has some “weight” $b$, so that we have $ax$ and $by$ (i.e. the monetary value, literal weight, cost, percent, etc. of the things).

These two things will often come with their respective totals:

3. The sum of the things: $x+y=T\,\text{, or the total quantity of things}$ (e.g. the total quantity of things purchased, sold, eated, etc.)
4. The sum of the “weights” of the things: $ax+by=R\,\text{, or the total weight of things}$ (e.g. the profit from selling fruits, the calories consumed from eating foods, etc.)

Based on the parameters of the questions, you could be looking for any combination of two of these things:

• the quantity of $x$
• the quantity of $y$
• the quantity of $a$
• the quantity of $b$
• the total of $x$ and $y$, $T$
• the total of $ax$ and $by$, $R$

However, since you will only likely ever be missing two of these quantities, you will solve these problems in the same way we have solved all the previous systems of equations in this section.

Example: 

Kate spent the day selling tickets for her school play. Adult tickets were priced at \$8, and child tickets were priced at \$6. She sold 27 tickets and collected \$192. How many adult tickets and how many child tickets did Kate sell?

So, we are selling tickets, which have two varieties: adult and child. So, let’s make our variables $A$ and $C$, so that we can keep everything straight in our minds. Thus,

$$A=\text{number of adult tickets sold}$$

$$C=\text{number of children tickets sold}$$

$$T=\text{total number of tickets sold}$$

$$A+C=T$$

We are told that the total number of tickets sold is 27, so

$$A+C=T\rightarrow{A+C=27}$$

What are the “weights” on $A$ and $C$? Well this would be the cost of $A$ and the cost of $C$, which we have been told are \$8 and \$6 respectively (i.e. we have the “weight” or literal value of $A$ as $8A$, and we have the “weight” or literal value of $C$ as $6C$). We are told that the total profit from ticket sales (i.e. the total of the “weights” of $A$ and $C$) is \$192. Thus, we can say that

$$8A+6C=192$$

Now, we have a system of two linear equations, which we may solve for $A$ and $C$, finding the number of adult and children tickets sold respectively. 

$$A+C=27$$

$$\underline{8A+6C=192}$$

$$ $$

$$-8\bigg(A+C=27\bigg)$$

$$\underline{8A+6C=192}$$

$$ $$

$$-8A-8C=-216$$

$$\underline{8A+6C=192}$$

$$0A-2C=-24$$

$$ $$

$$-2C=-24$$

$$C=12$$

$$ $$

$$A+C=17\rightarrow{A+12=27}$$

$$A=15$$

Thus, given the parameters of the situation, we must have sold 12 child tickets and 15 adult tickets.

Systems of nonlinear equations are usually best solved with substitution. They tend to yield more than a single solution when solutions exist. Often, they have a geometric interpretation as well. Consider the two equations given by $y^2 + x^2 = 4$ and $y = 4x^2 – 5$. The first equation represents a circle centered on the origin and the second equation represents a vertical parabola situated below this circle. Note, that we may reason that we are likely to see 0, 1, 2, or 4 solutions from this scenario.

Nonlinear equations tend to necessitate techniques of calculus or numerical solvers based on those techniques to obtain solutions.

Example 5: Solve $y^2 + x^2 = 4$ and $y = 4x^2 – 5$ using whatever method you decide.

The common approach is to use substitution from the less complicated equation and plug it into the more complicated equation. Here, it seems fairly straightforward to substitute equation 2 into equation 1 to yield equation 3.

$y^2 + x^2 = 4$ (equation 1)
$y = 4x^2 – 5$ (equation 2)
$(4x^2 – 5)^2 + x^2 = 4$ (equation 3)
$16x^4 -39x^2 + 21 = 0$ (equation 3*)

Equation 3* is quadratic in $x^2$ and we can make the substitution $w = x^2$ if we need to see this more obviously. Now, we apply the quadratic equation and obtain
$w = \frac{1}{32}(39 – \sqrt{2865}), \frac{1}{32}(39 + \sqrt{2865})$
We then take the positive and negative roots of both of these results. With a little work we can slightly simplify this to obtain:
$x = \pm \frac{1}{4} \sqrt{\frac{1}{2}(39 – \sqrt{177})}, \pm \frac{1}{4} \sqrt{\frac{1}{2}(39 + \sqrt{177})}$

Plugging these results into either equation 1 or 2 yields
$(-\frac{1}{4} \sqrt{\frac{1}{2}(39 – \sqrt{177})}, \frac{1}{8}(-1 – \sqrt{177}))$,
$(\frac{1}{4} \sqrt{\frac{1}{2}(39 – \sqrt{177})}, \frac{1}{8}(-1 – \sqrt{177}))$,
$(-\frac{1}{4} \sqrt{\frac{1}{2}(39 + \sqrt{177})}, \frac{1}{8}(-1 + \sqrt{177}))$,
$(\frac{1}{4} \sqrt{\frac{1}{2}(39 + \sqrt{177})}, \frac{1}{8}(-1 + \sqrt{177}))$

Example 6: Solve the system of equations given by $\frac{1}{4}y^2 + x^2 = 4$ and $y = log(x + 3)$

We can see from the graph that we have two solutions and can reason that we might have 0, 1, or 2 solutions from the outset. Again, we have an easy substitution, but even so we will need to use a numerical solver for one of the solutions.

$\frac{1}{4}(log(x+3))^2 + x^2 = 4$
$(log(x+3))^2 + 4x^2 = 16$
$(log(x+3))^2 + 4x^2 – 16 = 0$

By inspection, we can find x = -2 as a solution which yields the point of intersection (-2, 0). To find the second intersection we will employ a numerical solver. Though, by the second trimester, we will no longer need to do this using some technique of elementary calculus. We find the second solution is, to three decimal players, (1.838, 1.577).

1.

$$y\le{\frac{3}{2}x-2}$$

$$y\ge{13-x}$$

Consider the system of inequalities above. In the xy-plane, if a point with coordinates (c, d) lies in the solution set of the system of inequalities above, what is the minimum possible value of c?

2.

Portions of the graphs of two functions f(x) and g(x) are shown in the graph above. What is the x-coordinate of the point at which f(x) = g(x)?

3.

$$27x-7y=28$$

$$x-y={-1}$$

What is the solution $(x, y)$ to the system of equations above?

4.

$$12x+19y=15$$

$$ax-6y+{-5}$$

If the system of equations above has infinitely many solutions, what must be the value of $a$?

5. A store sells two different brands of rice, Brand A sells for \$4 per pound and Brand B sells for \$2 per pound. The store sold a total of 50 pounds of rice in a day and earned a total of \$130 from these sales. Which graph represents the correct solution of the number of pounds of each brand of rice sold at the store?