maths.free › Algebra › 5. Systems of Linear Equations › Solve Systems of Equations by Graphing

Solve Systems of Equations by Graphing

Determine whether an ordered pair is a solution of a system of equations

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations and Inequalities we learned how to solve linear equations with one variable. Remember that the solution of an equation is a value of the variable that makes a true statement when substituted into the equation.

Now we will work with systems of linear equations, two or more linear equations grouped together.

We will focus our work here on systems of two linear equations in two unknowns. Later, you may solve larger systems of equations.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

A linear equation in two variables, like 2x + y = 7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs (x, y) that make both equations true. These are called the solutions to a system of equations.

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

\[\{\begin{array}{l}3x-y=7 \\ x-2y=4\end{array}\]

Example

Try it.

Determine whether the ordered pair is a solution to the system: \(\{\begin{array}{l}x-y=-1 \\ 2x-y=-5\end{array}\)

ⓐ \((-2,-1)\) ⓑ \((-4,-3)\)

Solution

ⓐ

(–2, –1) does not make both equations true. (–2, –1) is not a solution.

ⓑ

(–4, –3) does make both equations true. (–4, –3) is a solution.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Solve a System of Linear Equations by Graphing

In this chapter we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown in :

For the first example of solving a system of linear equations in this section and in the next two sections, we will solve the same system of two linear equations. But we’ll use a different method in each section. After seeing the third method, you’ll decide which method was the most convenient way to solve this system.

How to Solve a System of Linear Equations by Graphing

Try it.

Solve the system by graphing: \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)

Solution

The steps to use to solve a system of linear equations by graphing are shown below.

Both equations in were given in slope–intercept form. This made it easy for us to quickly graph the lines. In the next example, we’ll first re-write the equations into slope–intercept form.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Determine the Number of Solutions of a Linear System

There will be times when we will want to know how many solutions there will be to a system of linear equations, but we might not actually have to find the solution. It will be helpful to determine this without graphing.

We have seen that two lines in the same plane must either intersect or are parallel. The systems of equations in through all had two intersecting lines. Each system had one solution.

A system with parallel lines, like , has no solution. What happened in ? The equations have coincident lines, and so the system had infinitely many solutions.

We’ll organize these results in below:

Parallel lines have the same slope but different y-intercepts. So, if we write both equations in a system of linear equations in slope–intercept form, we can see how many solutions there will be without graphing! Look at the system we solved in .

\[\begin{array}{llllllllll} & & & \ \{\ \begin{array}{lll}y & = & \frac{1}{2}x-3 \\ x-2y & = & 4\end{array} \\ \text{The first line is in slope-intercept form.} & & & \text{If we solve the second equation for}\ y,\ \text{we get} \\ y=\frac{1}{2}x-3 & & & \ \begin{array}{lll}x-2y & = & 4 \\ -2y & = & \text{-}x+4 \\ y & = & \frac{1}{2}x-2\end{array} \\ m=\frac{1}{2},b=-3 & & & m=\frac{1}{2},b=-2\end{array}\]

The two lines have the same slope but different y-intercepts. They are parallel lines.

shows how to determine the number of solutions of a linear system by looking at the slopes and intercepts.

\[\{\begin{array}{l}y=\frac{1}{2}x-3 \\ x-2y=4\end{array}\]\[y=\frac{1}{2}x-3\ y=\frac{1}{2}x-2\]

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Solve Applications of Systems of Equations by Graphing

We will use the same problem solving strategy we used in Math Models to set up and solve applications of systems of linear equations. We’ll modify the strategy slightly here to make it appropriate for systems of equations.

Step 5 is where we will use the method introduced in this section. We will graph the equations and find the solution.

Example

Try it.

Sondra is making 10 quarts of punch from fruit juice and club soda. The number of quarts of fruit juice is 4 times the number of quarts of club soda. How many quarts of fruit juice and how many quarts of club soda does Sondra need?

Solution

Step 1. Read the problem.

Step 2. Identify what we are looking for.

We are looking for the number of quarts of fruit juice and the number of quarts of club soda that Sondra will need.

Step 3. Name what we are looking for. Choose variables to represent those quantities.

Let \(f=\) number of quarts of fruit juice.
\(c=\) number of quarts of club soda

Step 4. Translate into a system of equations.

We now have the system. \(\{\begin{array}{l}f+c=10 \\ f=4c\end{array}\)

Step 5. Solve the system of equations using good algebra techniques.

The point of intersection (2, 8) is the solution. This means Sondra needs 2 quarts of club soda and 8 quarts of fruit juice.

Step 6. Check the answer in the problem and make sure it makes sense.

Does this make sense in the problem?

Yes, the number of quarts of fruit juice, 8 is 4 times the number of quarts of club soda, 2.

Yes, 10 quarts of punch is 8 quarts of fruit juice plus 2 quarts of club soda.

Step 7. Answer the question with a complete sentence.

Sondra needs 8 quarts of fruit juice and 2 quarts of soda.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

To solve a system of linear equations by graphing
1. Graph the first equation.
2. Graph the second equation on the same rectangular coordinate system.
3. Determine whether the lines intersect, are parallel, or are the same line.
4. Identify the solution to the system.
  If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system.
  If the lines are parallel, the system has no solution.
  If the lines are the same, the system has an infinite number of solutions.
5. Check the solution in both equations.
Determine the number of solutions from the graph of a linear system
Determine the number of solutions of a linear system by looking at the slopes and intercepts
Determine the number of solutions and how to classify a system of equations
Problem Solving Strategy for Systems of Linear Equations
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose variables to represent those quantities.
4. Translate into a system of equations.
5. Solve the system of equations using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Solve Systems of Equations by Graphing

Determine Whether an Ordered Pair is a Solution of a System of Equations. In the following exercises, determine if the following points are solutions to the given system of equations.

Try it.

\(\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\)

ⓐ \((3,1)\) ⓑ \((-3,4)\)

Solution

ⓐ yes ⓑ no

Try it.

\(\{\begin{array}{l}7x-4y=-1 \\ -3x-2y=1\end{array}\)

ⓐ \((1,2)\) ⓑ \((1,-2)\)

Try it.

\(\{\begin{array}{l}2x+y=5 \\ x+y=1\end{array}\)

ⓐ \((4,\text{-3})\) ⓑ \((2,0)\)

Solution

ⓐ yes ⓑ no

Try it.

\(\{\begin{array}{l}-3x+y=8 \\ -x+2y=-9\end{array}\)

ⓐ \((-5,-7)\) ⓑ \((-5,7)\)

Try it.

\(\{\begin{array}{l}x+y=2 \\ y=\frac{3}{4}x\end{array}\)

ⓐ \((\frac{8}{7},\frac{6}{7})\) ⓑ \((1,\frac{3}{4})\)

Solution

ⓐ yes ⓑ no

Try it.

\(\{\begin{array}{l}x+y=1 \\ y=\frac{2}{5}x\end{array}\)

ⓐ \((\frac{5}{7},\frac{2}{7})\) ⓑ \((5,2)\)

Try it.

\(\{\begin{array}{l}x+5y=10 \\ y=\frac{3}{5}x+1\end{array}\)

ⓐ \((-10,4)\) ⓑ \((\frac{5}{4},\frac{7}{4})\)

Solution

ⓐ no ⓑ yes

Try it.

\(\{\begin{array}{l}x+3y=9 \\ y=\frac{2}{3}x-2\end{array}\)

ⓐ \((-6,5)\) ⓑ \((5,\frac{4}{3})\)

Solve a System of Linear Equations by Graphing In the following exercises, solve the following systems of equations by graphing.

Try it.

\(\{\begin{array}{l}3x+y=-3 \\ 2x+3y=5\end{array}\)

Solution

\((-2,3)\)

Try it.

\(\{\begin{array}{l}\text{-}x+y=2 \\ 2x+y=-4\end{array}\)

Try it.

\(\{\begin{array}{l}-3x+y=-1 \\ 2x+y=4\end{array}\)

Solution

\((1,2)\)

Try it.

\(\{\begin{array}{l}-2x+3y=-3 \\ x+y=4\end{array}\)

Try it.

\(\{\begin{array}{l}y=x+2 \\ y=-2x+2\end{array}\)

Solution

\((0,2)\)

Try it.

\(\{\begin{array}{l}y=x-2 \\ y=-3x+2\end{array}\)

Try it.

\(\{\begin{array}{l}y=\frac{3}{2}x+1 \\ y=-\frac{1}{2}x+5\end{array}\)

Solution

\((2,4)\)

Try it.

\(\{\begin{array}{l}y=\frac{2}{3}x-2 \\ y=-\frac{1}{3}x-5\end{array}\)

Try it.

\(\{\begin{array}{l}-x+y=-3 \\ 4x+4y=4\end{array}\)

Solution

\((2,-1)\)

Try it.

\(\{\begin{array}{l}x-y=3 \\ 2x-y=4\end{array}\)

Try it.

\(\{\begin{array}{l}-3x+y=-1 \\ 2x+y=4\end{array}\)

Solution

\((1,2)\)

Try it.

\(\{\begin{array}{l}-3x+y=-2 \\ 4x-2y=6\end{array}\)

Try it.

\(\{\begin{array}{l}x+y=5 \\ 2x-y=4\end{array}\)

Solution

\((3,2)\)

Try it.

\(\{\begin{array}{l}x-y=2 \\ 2x-y=6\end{array}\)

Try it.

\(\{\begin{array}{l}x+y=2 \\ x-y=0\end{array}\)

Solution

\((1,1)\)

Try it.

\(\{\begin{array}{l}x+y=6 \\ x-y=-8\end{array}\)

Try it.

\(\{\begin{array}{l}x+y=-5 \\ x-y=3\end{array}\)

Solution

\((-1,-4)\)

Try it.

\(\{\begin{array}{l}x+y=4 \\ x-y=0\end{array}\)

Try it.

\(\{\begin{array}{l}x+y=-4 \\ -x+2y=-2\end{array}\)

Solution

\((-2,-2)\)

Try it.

\(\{\begin{array}{l}-x+3y=3 \\ x+3y=3\end{array}\)

Try it.

\(\{\begin{array}{l}-2x+3y=3 \\ x+3y=12\end{array}\)

Solution

\((3,3)\)

Try it.

\(\{\begin{array}{l}2x-y=4 \\ 2x+3y=12\end{array}\)

Try it.

\(\{\begin{array}{l}2x+3y=6 \\ y=-2\end{array}\)

Solution

\((6,-2)\)

Try it.

\(\{\begin{array}{l}-2x+y=2 \\ y=4\end{array}\)

Try it.

\(\{\begin{array}{l}x-3y=-3 \\ y=2\end{array}\)

Solution

\((3,2)\)

Try it.

\(\{\begin{array}{l}2x-2y=8 \\ y=-3\end{array}\)

Try it.

\(\{\begin{array}{l}2x-y=-1 \\ x=1\end{array}\)

Solution

\((1,3)\)

Try it.

\(\{\begin{array}{l}x+2y=2 \\ x=-2\end{array}\)

Try it.

\(\{\begin{array}{l}x-3y=-6 \\ x=-3\end{array}\)

Solution

\((-3,1)\)

Try it.

\(\{\begin{array}{l}x+y=4 \\ x=1\end{array}\)

Try it.

\(\{\begin{array}{l}4x-3y=8 \\ 8x-6y=14\end{array}\)

Solution

no solution

Try it.

\(\{\begin{array}{l}x+3y=4 \\ -2x-6y=3\end{array}\)

Try it.

\(\{\begin{array}{l}-2x+4y=4 \\ y=\frac{1}{2}x\end{array}\)

Solution

no solution

Try it.

\(\{\begin{array}{l}3x+5y=10 \\ y=-\frac{3}{5}x+1\end{array}\)

Try it.

\(\{\begin{array}{l}x=-3y+4 \\ 2x+6y=8\end{array}\)

Solution

infinitely many solutions

Try it.

\(\{\begin{array}{l}4x=3y+7 \\ 8x-6y=14\end{array}\)

Try it.

\(\{\begin{array}{l}2x+y=6 \\ -8x-4y=-24\end{array}\)

Solution

infinitely many solutions

Try it.

\(\{\begin{array}{l}5x+2y=7 \\ -10x-4y=-14\end{array}\)

Try it.

\(\{\begin{array}{l}x+3y=-6 \\ 4y=-\frac{4}{3}x-8\end{array}\)

Solution

infinitely many solutions

Try it.

\(\{\begin{array}{l}\text{-}x+2y=-6 \\ y=-\frac{1}{2}x-1\end{array}\)

Try it.

\(\{\begin{array}{l}-3x+2y=-2 \\ y=\text{-}x+4\end{array}\)

Solution

\((2,2)\)

Try it.

\(\{\begin{array}{l}-x+2y=-2 \\ y=\text{-}x-1\end{array}\)

Determine the Number of Solutions of a Linear System Without graphing the following systems of equations, determine the number of solutions and then classify the system of equations.

Try it.

\(\{\begin{array}{l}y=\frac{2}{3}x+1 \\ -2x+3y=5\end{array}\)

Solution

0 solutions

Try it.

\(\{\begin{array}{l}y=\frac{1}{3}x+2 \\ x-3y=9\end{array}\)

Try it.

\(\{\begin{array}{l}y=-2x+1 \\ 4x+2y=8\end{array}\)

Solution

0 solutions

Try it.

\(\{\begin{array}{l}y=3x+4 \\ 9x-3y=18\end{array}\)

Try it.

\(\{\begin{array}{l}y=\frac{2}{3}x+1 \\ 2x-3y=7\end{array}\)

Solution

no solutions, inconsistent, independent

Try it.

\(\{\begin{array}{l}3x+4y=12 \\ y=-3x-1\end{array}\)

Try it.

\(\{\begin{array}{l}4x+2y=10 \\ 4x-2y=-6\end{array}\)

Solution

consistent, 1 solution

Try it.

\(\{\begin{array}{l}5x+3y=4 \\ 2x-3y=5\end{array}\)

Try it.

\(\{\begin{array}{l}y=-\frac{1}{2}x+5 \\ x+2y=10\end{array}\)

Solution

infinitely many solutions

Try it.

\(\{\begin{array}{l}y=x+1 \\ \text{-}x+y=1\end{array}\)

Try it.

\(\{\begin{array}{l}y=2x+3 \\ 2x-y=-3\end{array}\)

Solution

infinitely many solutions

Try it.

\(\{\begin{array}{l}5x-2y=10 \\ y=\frac{5}{2}x-5\end{array}\)

Solve Applications of Systems of Equations by Graphing In the following exercises, solve.

Try it.

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Solution

Molly needs 16 ounces of strawberry juice and 48 ounces of water.

Try it.

Jamal is making a snack mix that contains only pretzels and nuts. For every ounce of nuts, he will use 2 ounces of pretzels. How many ounces of pretzels and how many ounces of nuts does he need to make 45 ounces of snack mix?

Try it.

Enrique is making a party mix that contains raisins and nuts. For each ounce of nuts, he uses twice the amount of raisins. How many ounces of nuts and how many ounces of raisins does he need to make 24 ounces of party mix?

Solution

Enrique needs 8 ounces of nuts and 16 ounces of water.

Try it.

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For the equation \(y=\frac{2}{3}x-4\)
ⓐ is \((6,0)\) a solution? ⓑ is \((-3,-2)\) a solution?
If you missed this problem, review .

Jawaabta muuji

(a) yes (b) no
Find the slope and y-intercept of the line \(3x-y=12\).
If you missed this problem, review .

Jawaabta muuji

\(m=3;\) \(b=-12\)
Find the x- and y-intercepts of the line \(2x-3y=12\).
If you missed this problem, review .

Jawaabta muuji

\((6,0),\ (0,-4)\)
Determine whether the ordered pair is a solution to the system: \(\{\begin{array}{l}x-y=-1 \\ 2x-y=-5\end{array}\)
ⓐ \((-2,-1)\) ⓑ \((-4,-3)\)
Jawaabta muuji
1. ⓐ
  
  (–2, –1) does not make both equations true. (–2, –1) is not a solution.
  
  ⓑ
  
  (–4, –3) does make both equations true. (–4, –3) is a solution.
Determine whether the ordered pair is a solution to the system: \(\{\begin{array}{l}3x+y=0 \\ x+2y=-5\end{array}.\)
ⓐ \((1,-3)\) ⓑ \((0,0)\)

Jawaabta muuji

ⓐ yes ⓑ no
Determine whether the ordered pair is a solution to the system: \(\{\begin{array}{l}x-3y=-8 \\ -3x-y=4\end{array}.\)
ⓐ \((2,-2)\) ⓑ \((-2,2)\)

Jawaabta muuji

ⓐ no ⓑ yes
Solve the system by graphing: \(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}.\)
Solve the system by graphing: \(\{\begin{array}{l}x-3y=-3 \\ x+y=5\end{array}.\)

Jawaabta muuji

\((3,2)\)
Solve the system by graphing: \(\{\begin{array}{l}-x+y=1 \\ 3x+2y=12\end{array}.\)

Jawaabta muuji

\((2,3)\)

Solve the system by graphing: \(\{\begin{array}{l}y=2x+1 \\ y=4x-1\end{array}.\)

Jawaabta muuji

Both of the equations in this system are in slope-intercept form, so we will use their slopes and y-intercepts to graph them. \(\{\begin{array}{l}y=2x+1 \\ y=4x-1\end{array}\)

Find the slope and y-intercept of the first equation.
Find the slope and y-intercept of the first equation.
Graph the two lines.
Determine the point of intersection.	The lines intersect at (1, 3).

Check the solution in both equations.	\(\begin{array}{llllllllllllllll}\begin{array}{lll}y & = & 2x+1 \\ 3 & \overset{?}{=} & 2\cdot 1+1 \\ 3 & = & 3\ ✓\end{array} & & & \begin{array}{lll}y & = & 4x-1 \\ 3 & \overset{?}{=} & 4\cdot 1-1 \\ 3 & = & 3\ ✓\end{array}\end{array}\)
	The solution is (1, 3).

Solve each system by graphing: \(\{\begin{array}{l}y=2x+2 \\ y=\text{-}x-4\end{array}.\)

Jawaabta muuji

\((-2,-2)\)
Solve each system by graphing: \(\{\begin{array}{l}y=3x+3 \\ y=\text{-}x+7\end{array}.\)

Jawaabta muuji

\((1,6)\)

Solve the system by graphing: \(\{\begin{array}{l}3x+y=-1 \\ 2x+y=0\end{array}.\)

Jawaabta muuji

We’ll solve both of these equations for \(y\) so that we can easily graph them using their slopes and y-intercepts. \(\{\begin{array}{l}3x+y=-1 \\ 2x+y=0\end{array}\)

Solve the first equation for y. Find the slope and y-intercept. Solve the second equation for y. Find the slope and y-intercept.	\(\begin{array}{lllllllllllllllll}\begin{array}{lll}3x+y & = & -1 \\ y & = & -3x-1 \\ \\ m & = & -3 \\ b & = & -1 \\ \\ \\ 2x+y & = & 0 \\ y & = & -2x \\ \\ m & = & -2 \\ b & = & 0 \\ \end{array}\end{array}\)
Graph the lines.
Determine the point of intersection.	The lines intersect at (−1, 2).
Check the solution in both equations.	\(\begin{array}{llllllllllllllll}\begin{array}{lll}3x+y & = & -1 \\ 3(-1)+2 & \overset{?}{=} & -1 \\ -1 & = & -1\ ✓\end{array} & & & \begin{array}{lll}2x+y & = & 0 \\ 2(-1)+2 & \overset{?}{=} & 0 \\ 0 & = & 0\ ✓\end{array}\end{array}\)
	The solution is (−1, 2).

Solve each system by graphing: \(\{\begin{array}{l}-x+y=1 \\ 2x+y=10\end{array}.\)

Jawaabta muuji

\((3,4)\)
Solve each system by graphing: \(\{\begin{array}{l}2x+y=6 \\ x+y=1\end{array}.\)

Jawaabta muuji

\((5,-4)\)

Solve the system by graphing: \(\{\begin{array}{l}x+y=2 \\ x-y=4\end{array}.\)

Jawaabta muuji

We will find the x- and y-intercepts of both equations and use them to graph the lines.


To find the intercepts, let x = 0 and solve for y, then let y = 0 and solve for x.	\(\begin{array}{llllllllllllllll}\begin{array}{lll}x+y & = & 2 \\ 0+y & = & 2 \\ y & = & 2\end{array} & & & \begin{array}{lll}x+y & = & 2 \\ x+0 & = & 2 \\ x & = & 2\end{array}\end{array}\)

To find the intercepts, let x = 0 then let y = 0.	\(\begin{array}{llllllllllllllllll}\begin{array}{lll}x-y & = & 4 \\ 0-y & = & 4 \\ -y & = & 4 \\ y & = & -4\end{array} & & & \begin{array}{lll}x-y & = & 4 \\ x-0 & = & 4 \\ x & = & 4 \\ \\ \\ \end{array}\end{array}\)
Graph the line.
Determine the point of intersection.	The lines intersect at (3, −1).
Check the solution in both equations.	\(\begin{array}{llllllll}x+y & = & 2 & & & x-y & = & 4 \\ 3+(-1) & \overset{?}{=} & 2 & & & 3-(-1) & \overset{?}{=} & 4 \\ 2 & = & 2✓ & & & 4 & = & 4✓\end{array}\) The solution is (3, −1).

Solve each system by graphing: \(\{\begin{array}{l}x+y=6 \\ x-y=2\end{array}.\)

Jawaabta muuji

\((4,2)\)
Solve each system by graphing: \(\{\begin{array}{l}x+y=2 \\ x-y=-8\end{array}.\)

Jawaabta muuji

\((-3,5)\)

Solve the system by graphing: \(\{\begin{array}{l}y=6 \\ 2x+3y=12\end{array}.\)

Jawaabta muuji


We know the first equation represents a horizontal line whose y-intercept is 6.
The second equation is most conveniently graphed using intercepts.
To find the intercepts, let x = 0 and then y = 0.
Graph the lines.
Determine the point of intersection.	The lines intersect at (−3, 6).
Check the solution to both equations.	\(\begin{array}{llllllll}y & = & 6 & & & 2x+3y & = & 12 \\ 6 & \overset{?}{=} & 6✓ & & & 2(-3)+3(6) & \overset{?}{=} & 12 \\ & & & & & -6+18 & \overset{?}{=} & 12 \\ & & & & & 12 & = & 12✓\end{array}\)
	The solution is (−3, 6).

Solve each system by graphing: \(\{\begin{array}{l}y=-1 \\ x+3y=6\end{array}.\)

Jawaabta muuji

\((9,-1)\)
Solve each system by graphing: \(\{\begin{array}{l}x=4 \\ 3x-2y=24\end{array}.\)

Jawaabta muuji

\((4,-6)\)

Solve the system by graphing: \(\{\begin{array}{l}y=\frac{1}{2}x-3 \\ x-2y=4\end{array}.\)

Jawaabta muuji


To graph the first equation, we will use its slope and y-intercept.


To graph the second equation, we will use the intercepts.

Graph the lines.
Determine the point of intersection.	The lines are parallel.
	Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system.

Solve each system by graphing: \(\{\begin{array}{l}y=-\frac{1}{4}x+2 \\ x+4y=-8\end{array}.\)

Jawaabta muuji

no solution
Solve each system by graphing: \(\{\begin{array}{l}y=3x-1 \\ 6x-2y=6\end{array}.\)

Jawaabta muuji

no solution

Solve the system by graphing: \(\{\begin{array}{l}y=2x-3 \\ -6x+3y=-9\end{array}.\)

Jawaabta muuji


Find the slope and y-intercept of the first equation.
Find the intercepts of the second equation.

Graph the lines.
Determine the point of intersection.	The lines are the same!
	Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true.
	There are infinitely many solutions to this system.

Solve each system by graphing: \(\{\begin{array}{l}y=-3x-6 \\ 6x+2y=-12\end{array}.\)

Jawaabta muuji

infinitely many solutions
Solve each system by graphing: \(\{\begin{array}{l}y=\frac{1}{2}x-4 \\ 2x-4y=16\end{array}.\)

Jawaabta muuji

infinitely many solutions

Without graphing, determine the number of solutions and then classify the system of equations: \(\{\begin{array}{l}y=3x-1 \\ 6x-2y=12\end{array}.\)

Jawaabta muuji

We will compare the slopes and intercepts of the two lines.	\(\{\begin{array}{l}y=3x-1 \\ 6x-2y=12\end{array}.\)
The first equation is already in slope-intercept form.	\(y=3x-1\)
Write the second equation in slope-intercept form.	\(\begin{array}{lll}6x-2y & = & 12 \\ -2y & = & -6x+12 \\ \frac{-2y}{-2} & = & \frac{-6x+12}{-2} \\ y & = & 3x-6\end{array}\)
Find the slope and intercept of each line.	\(\begin{array}{lllllllll}y & = & 3x-1 & & & & y & = & 3x-6 \\ m & = & 3 & & & & m & = & 3 \\ b & = & -1 & & & & b & = & -6\end{array}\)
	Since the slopes are the same and \(y\)-intercepts are different, the lines are parallel.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

Without graphing, determine the number of solutions and then classify the system of equations.
\(\{\begin{array}{l}y=-2x-4 \\ 4x+2y=9\end{array}\)

Jawaabta muuji

no solution, inconsistent, independent
Without graphing, determine the number of solutions and then classify the system of equations.
\(\{\begin{array}{l}y=\frac{1}{3}x-5 \\ x-3y=6\end{array}\)

Jawaabta muuji

no solution, inconsistent, independent

Without graphing, determine the number of solutions and then classify the system of equations: \(\{\begin{array}{l}2x+y=-3 \\ x-5y=5\end{array}.\)

Jawaabta muuji

We will compare the slope and intercepts of the two lines.	\(\{\begin{array}{l}2x+y=-3 \\ x-5y=5\end{array}\)
Write both equations in slope-intercept form.	\(\begin{array}{lll}2x+y & = & -3 \\ y & = & -2x-3\end{array}\)	\(\begin{array}{lll}x-5y=5 & = & 5 \\ -5y & = & -x+5 \\ \frac{-5y}{-5} & = & \frac{-x+5}{-5} \\ y & = & \frac{1}{5}x-1\end{array}\)
Find the slope and intercept of each line.	\(\begin{array}{lll}y & = & -2x-3 \\ m & = & -2 \\ b & = & -3\end{array}\)	\(\begin{array}{lll}y & = & \frac{1}{5}x-1 \\ m & = & \frac{1}{5} \\ b & = & -1\end{array}\)
	Since the slopes are different, the lines intersect.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Without graphing, determine the number of solutions and then classify the system of equations.
\(\{\begin{array}{l}3x+2y=2 \\ 2x+y=1\end{array}\)

Jawaabta muuji

one solution, consistent, independent
Without graphing, determine the number of solutions and then classify the system of equations.
\(\{\begin{array}{l}x+4y=12 \\ -x+y=3\end{array}\)

Jawaabta muuji

one solution, consistent, independent

Without graphing, determine the number of solutions and then classify the system of equations. \(\{\begin{array}{l}3x-2y=4 \\ y=\frac{3}{2}x-2\end{array}\)

Jawaabta muuji

We will compare the slopes and intercepts of the two lines.	\(\{\begin{array}{l}3x-2y=4 \\ y=\frac{3}{2}x-2\end{array}\)
Write the first equation in slope-intercept form.	\(\begin{array}{lll}3x-2y & = & 4 \\ -2y & = & -3x+4 \\ \frac{-2y}{-2} & = & \frac{-3x+4}{-2} \\ y & = & \frac{3}{2}x-2\end{array}\)
The second equation is already in slope-intercept form.	\(\begin{array}{l}y=\frac{3}{2}x-2\end{array}\)
	Since the slopes are the same, they have the same slope and same \(y\)-intercept and so the lines are coincident.

A system of equations whose graphs are coincident lines has infinitely many solutions and is consistent and dependent.

Without graphing, determine the number of solutions and then classify the system of equations.
\(\{\begin{array}{l}4x-5y=20 \\ y=\frac{4}{5}x-4\end{array}\)

Jawaabta muuji

infinitely many solutions, consistent, dependent
Without graphing, determine the number of solutions and then classify the system of equations.
\(\{\begin{array}{l}-2x-4y=8 \\ y=-\frac{1}{2}x-2\end{array}\)

Jawaabta muuji

infinitely many solutions, consistent, dependent
Sondra is making 10 quarts of punch from fruit juice and club soda. The number of quarts of fruit juice is 4 times the number of quarts of club soda. How many quarts of fruit juice and how many quarts of club soda does Sondra need?

Jawaabta muuji

Step 1. Read the problem.
Step 2. Identify what we are looking for.
We are looking for the number of quarts of fruit juice and the number of quarts of club soda that Sondra will need.
Step 3. Name what we are looking for. Choose variables to represent those quantities.
Let \(f=\) number of quarts of fruit juice.
\(c=\) number of quarts of club soda
Step 4. Translate into a system of equations.

We now have the system. \(\{\begin{array}{l}f+c=10 \\ f=4c\end{array}\)
Step 5. Solve the system of equations using good algebra techniques.

The point of intersection (2, 8) is the solution. This means Sondra needs 2 quarts of club soda and 8 quarts of fruit juice.
Step 6. Check the answer in the problem and make sure it makes sense.
Does this make sense in the problem?
Yes, the number of quarts of fruit juice, 8 is 4 times the number of quarts of club soda, 2.
Yes, 10 quarts of punch is 8 quarts of fruit juice plus 2 quarts of club soda.
Step 7. Answer the question with a complete sentence.
Sondra needs 8 quarts of fruit juice and 2 quarts of soda.
Manny is making 12 quarts of orange juice from concentrate and water. The number of quarts of water is 3 times the number of quarts of concentrate. How many quarts of concentrate and how many quarts of water does Manny need?

Jawaabta muuji

Manny needs 3 quarts juice concentrate and 9 quarts water.
Alisha is making an 18 ounce coffee beverage that is made from brewed coffee and milk. The number of ounces of brewed coffee is 5 times greater than the number of ounces of milk. How many ounces of coffee and how many ounces of milk does Alisha need?

Jawaabta muuji

Alisha needs 15 ounces of coffee and 3 ounces of milk.
\(\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\)
ⓐ \((3,1)\) ⓑ \((-3,4)\)

Jawaabta muuji

ⓐ yes ⓑ no

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Systems of Equations by Graphing

Determine whether an ordered pair is a solution of a system of equations
Solve a system of linear equations by graphing
Determine the number of solutions of linear system
Solve applications of systems of equations by graphing
Graph the first equation.
Graph the second equation on the same rectangular coordinate system.
Determine whether the lines intersect, are parallel, or are the same line.
Identify the solution to the system.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Ku day inaad ku

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

In ka badan Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value