How to solve system of equations using substitution

Play with our fun little avatar builder to create and customize your own avatar on StudyPug. Choose your face, eye colour, hair colour and style, and background. Unlock more options the more you use StudyPug.

Table of Contents Show

Solve a System of Equations by Substitution
Solve Applications of Systems of Equations by Substitution
Key Concepts
How do you solve by substitution step by step?
What are the three steps when solving system of equations by substitution?
How do you solve two equations with substitution?
Which system of equations would be best solved using substitution?

Because it is used in such topics as nonlinear systems, linear algebra, computer programming, and so much more.

And the greatest thing about solving systems by substitution is that it’s easy to use!

The method of substitution involves three steps:

Solve one equation for one of the variables.
Substitute (plug-in) this expression into the other equation and solve.
Resubstitute the value into the original equation to find the corresponding variable.

Now at first glance, this may seem complicated, but I’ve got some helpful tricks for keeping things straight. In fact, we’re going to make a sort of circular circuit that helps to provide organization and efficiency to our method.

Using the Substitution Method to Solve

Remember, our goal when solving any system is to find the point of intersection. As we saw in our lesson titled the graphing method, we saw that some systems do not have solutions because they don’t intersect, and others coincide, which provides infinitely many solutions.

So when we solve systems by substitution, we will need to be on the lookout for these types of scenarios. If they are parallel and don’t intersect, then we are going to end up with an invalid answer, or as Purple Math calls it, a “garbage” result.

Together we will look at 11 examples of solving linear systems using the substitution method, and learn how to employ this technique for systems of two, three and even four equations.

In two variables (x and y) , the graph of a system of two equations is a pair of lines in the plane.

There are three possibilities:

The lines intersect at zero points. (The lines are parallel.)
The lines intersect at exactly one point. (Most cases.)
The lines intersect at infinitely many points. (The two equations represent the same line.)

How to Solve a System Using The Substitution Method

Step 1 : First, solve one linear equation for y in terms of x .
Step 2 : Then substitute that expression for y in the other linear equation. You'll get an equation in x .
Step 3 : Solve this, and you have the x -coordinate of the intersection.
Step 4 : Then plug in x to either equation to find the corresponding y -coordinate.

Note 1 : If it's easier, you can start by solving an equation for x in terms of y , also – same difference!

Example:

Solve the system {3x+2y=167x+y=19

Solve the second equation for y .

y=19−7x

Substitute 19−7x for y in the first equation and solve for x .

3x+2(19−7x)=163x+38−14x=16−11x=−22x=2

Substitute 2 for x in y=19−7x and solve for y .

y=19−7(2)y=5

Note 2 : If the lines are parallel, your x -terms will cancel in step 2 , and you will get an impossible equation, something like 0=3 .

Note 3 : If the two equations represent the same line, everything will cancel in step 2 , and you will get a redundant equation, 0=0 .

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Learning Objectives
Note
Solve a System of Equations by Substitution
1. Exercise $\PageIndex{1}$: How to Solve a System of Equations by Substitution
2. Exercise $\PageIndex{2}$
3. Exercise $\PageIndex{3}$
4. SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.
5. Exercise $\PageIndex{4}$
6. Exercise $\PageIndex{5}$
7. Exercise $\PageIndex{6}$
8. Exercise $\PageIndex{7}$
9. Exercise $\PageIndex{8}$
10. Exercise $\PageIndex{9}$
11. Exercise $\PageIndex{10}$
12. Exercise $\PageIndex{11}$
13. Exercise $\PageIndex{12}$
14. Exercise $\PageIndex{13}$
15. Exercise $\PageIndex{14}$
16. Exercise $\PageIndex{15}$
17. Exercise $\PageIndex{16}$
18. Exercise $\PageIndex{17}$
19. Exercise $\PageIndex{18}$
20. Exercise $\PageIndex{19}$
21. Exercise $\PageIndex{20}$
22. Exercise $\PageIndex{21}$
23. Exercise $\PageIndex{22}$
24. Exercise $\PageIndex{23}$
25. Exercise $\PageIndex{24}$
Solve Applications of Systems of Equations by Substitution
1. HOW TO USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.
2. Exercise $\PageIndex{25}$
3. Exercise $\PageIndex{26}$
4. Exercise $\PageIndex{27}$
5. Exercise $\PageIndex{28}$
6. Exercise $\PageIndex{29}$
7. Exercise $\PageIndex{30}$
8. Exercise $\PageIndex{31}$
9. Exercise $\PageIndex{32}$
10. Exercise $\PageIndex{33}$
11. Exercise $\PageIndex{34}$
12. Exercise $\PageIndex{35}$
13. Exercise $\PageIndex{36}$
14. Note
Key Concepts

Learning Objectives

By the end of this section, you will be able to:

Solve a system of equations by substitution
Solve applications of systems of equations by substitution

Note

Before you get started, take this readiness quiz.

Simplify −5(3−x).
If you missed this problem, review Exercise 1.10.43.
Simplify 4−2(n+5).
If you missed this problem, review Exercise 1.10.41.
Solve for y. 8y−8=32−2y
If you missed this problem, review Exercise 2.3.22.
Solve for x. 3x−9y=−3
If you missed this problem, review Exercise 2.6.22.

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

$\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.$

We will first solve one of the equations for either x or y. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Exercise $\PageIndex{1}$.

Exercise $\PageIndex{1}$: How to Solve a System of Equations by Substitution

Solve the system by substitution. $\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.$

Answer

Exercise $\PageIndex{2}$

Solve the system by substitution. $\left\{\begin{array}{l}{-2 x+y=-11} \\ {x+3 y=9}\end{array}\right.$

Answer

(6,1)

Exercise $\PageIndex{3}$

Solve the system by substitution. $\left\{\begin{array}{l}{x+3 y=10} \\ {4 x+y=18}\end{array}\right.$

Answer

(4,2)

SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.

Solve one of the equations for either variable.
Substitute the expression from Step 1 into the other equation.
Solve the resulting equation.
Substitute the solution in Step 3 into one of the original equations to find the other variable.
Write the solution as an ordered pair.
Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Exercise $\PageIndex{4}$.

Exercise $\PageIndex{4}$

Solve the system by substitution. $\left\{\begin{array}{l}{x+y=-1} \\ {y=x+5}\end{array}\right.$

Answer

The second equation is already solved for y. We will substitute the expression in place of y in the first equation.

The second equation is already solved for y.
We will substitute into the first equation. Replace the y with x + 5.

Solve the resulting equation for x.

Substitute x = −3 into y = x + 5 to find y.

The ordered pair is (−3, 2).

Check the ordered pair in both equations:

$\begin{array} {rllrll} x+y &=&-1 & y&=&x+5\\-3+2 &\stackrel{?}{=}&-1 &2& \stackrel{?}{=} & -3 + 5\\-1 &=&-1\checkmark &2 &=&2\checkmark \end{array}$ The solution is (−3, 2).

Exercise $\PageIndex{5}$

Solve the system by substitution. $\left\{\begin{array}{l}{x+y=6} \\ {y=3 x-2}\end{array}\right.$

Answer

(2,4)

Exercise $\PageIndex{6}$

Solve the system by substitution. $\left\{\begin{array}{l}{2 x-y=1} \\ {y=-3 x-6}\end{array}\right.$

Answer

(−1,−3)

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y.

Exercise $\PageIndex{7}$

Solve the system by substitution. $\left\{\begin{array}{l}{3 x+y=5} \\ {2 x+4 y=-10}\end{array}\right.$

Answer

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Solve for y.

Substitute into the other equation.

Replace the y with −3x + 5.

Solve the resulting equation for x.

Substitute x = 3 into 3x + y = 5 to find y.

The ordered pair is (3, −4).

Check the ordered pair in both equations:

$\begin{array} {rllrll} 3x+y &=&5 & 2x+4y&=&-10\\3\cdot3+(-4) &\stackrel{?}{=}&5 &2\cdot3 + 4(-4)& \stackrel{?}{=} & -10\\9-4&\stackrel{?}{=}&5 &6-16& \stackrel{?}{=} & -10\\5 &=&5\checkmark &-10&=&-10\checkmark \end{array}$

The solution is (3, −4).

Exercise $\PageIndex{8}$

Solve the system by substitution. $\left\{\begin{array}{l}{4 x+y=2} \\ {3 x+2 y=-1}\end{array}\right.$

Answer

(1,−2)

Exercise $\PageIndex{9}$

Solve the system by substitution. $\left\{\begin{array}{l}{-x+y=4} \\ {4 x-y=2}\end{array}\right.$

Answer

(2,6)

In Exercise $\PageIndex{7}$ it was easiest to solve for y in the first equation because it had a coefficient of 1. In Exercise $\PageIndex{10}$ it will be easier to solve for x.

Exercise $\PageIndex{10}$

Solve the system by substitution. $\left\{\begin{array}{l}{x-2 y=-2} \\ {3 x+2 y=34}\end{array}\right.$

Answer

We will solve the first equation for xx and then substitute the expression into the second equation.

Solve for x.

Substitute into the other equation.

Replace the x with 2y − 2.

Solve the resulting equation for y.

Substitute y = 5 into x − 2y = −2 to find x.

The ordered pair is (8, 5). Check the ordered pair in both equations:

$\begin{array} {rllrll} x-2y &=&-2 & 3x+2y&=&34\\8-2\cdot 5 &\stackrel{?}{=}&-2 &3\cdot8 + 2\cdot5& \stackrel{?}{=} & 34\\8-10&\stackrel{?}{=}&-2 &24+10& \stackrel{?}{=} & 34\\-2 &=&-2\checkmark &34&=&34\checkmark \end{array}$ The solution is (8, 5).

Exercise $\PageIndex{11}$

Solve the system by substitution. $\left\{\begin{array}{l}{x-5 y=13} \\ {4 x-3 y=1}\end{array}\right.$

Answer

(−2,−3)

Exercise $\PageIndex{12}$

Solve the system by substitution. $\left\{\begin{array}{l}{x-6 y=-6} \\ {2 x-4 y=4}\end{array}\right.$

Answer

(6,2)

When both equations are already solved for the same variable, it is easy to substitute!

Exercise $\PageIndex{13}$

Solve the system by substitution. $\left\{\begin{array}{l}{y=-2 x+5} \\ {y=\frac{1}{2} x}\end{array}\right.$

Answer

Since both equations are solved for y, we can substitute one into the other.

Substitute $\frac{1}{2}x$ for y in the first equation.

Replace the y with $\frac{1}{2}x$

Solve the resulting equation. Start
by clearing the fraction.

Solve for x.

Substitute x = 2 into $y = \frac{1}{2}x$ to find y.

The ordered pair is (2,1). Check the ordered pair in both equations:

$\begin{array} {rllrll} y &=&\frac{1}{2}x & y&=&-2x+5\\1 &\stackrel{?}{=}&\frac{1}{2}\cdot2 &1& \stackrel{?}{=} & -2\cdot2+5\\1 &=&1\checkmark &1 &=&-4+5\\ &&&1&=&1\checkmark \end{array}$ The solution is (2,1).

Exercise $\PageIndex{14}$

Solve the system by substitution. $\left\{\begin{array}{l}{y=3 x-16} \\ {y=\frac{1}{3} x}\end{array}\right.$

Answer

(6,2)

Exercise $\PageIndex{15}$

Solve the system by substitution. $\left\{\begin{array}{l}{y=-x+10} \\ {y=\frac{1}{4} x}\end{array}\right.$

Answer

(8,2)

Be very careful with the signs in the next example.

Exercise $\PageIndex{16}$

Solve the system by substitution. $\left\{\begin{array}{l}{4 x+2 y=4} \\ {6 x-y=8}\end{array}\right.$

Answer

We need to solve one equation for one variable. We will solve the first equation for y.

Solve the first equation for y.

Substitute −2x + 2 for y in the second equation.

Replace the y with −2x + 2.

Solve the equation for x.

Substitute $x = \frac{5}{4}$ into 4x + 2y = 4 to find y.

The ordered pair is $(\frac{5}{4},−\frac{1}{2})$. Check the ordered pair in both equations.

$\begin{array} {rllrll} 4x+2y &=&4& 6x-y&=&8\\4(\frac{5}{4}) +2(-\frac{1}{2})&\stackrel{?}{=}&4 &6(\frac{5}{4}) - (-\frac{1}{2})& \stackrel{?}{=} & 8\\5-1&\stackrel{?}{=}&4 &\frac{15}{4} - (-\frac{1}{2}) &\stackrel{?}{=} & 8\\4 &=&4\checkmark &\frac{16}{2} &\stackrel{?}{=}&8\\ &&&8&=&8\checkmark \end{array}$ The solution is (54,−12).

Exercise $\PageIndex{17}$

Solve the system by substitution. $\left\{\begin{array}{l}{x-4 y=-4} \\ {-3 x+4 y=0}\end{array}\right.$

Answer

$(2,\frac{3}{2})$

Exercise $\PageIndex{18}$

Solve the system by substitution. $\left\{\begin{array}{l}{4 x-y=0} \\ {2 x-3 y=5}\end{array}\right.$

Answer

$(−\frac{1}{2},−2)$

In Example, it will take a little more work to solve one equation for x or y.

Exercise $\PageIndex{19}$

Solve the system by substitution. $\left\{\begin{array}{l}{4 x-3 y=6} \\ {15 y-20 x=-30}\end{array}\right.$

Answer

We need to solve one equation for one variable. We will solve the first equation for x.

Solve the first equation for x.

Substitute $\frac{3}{4} y+\frac{3}{2}$ for x in the second equation.

Replace the x with $\frac{3}{4} y+\frac{3}{2}$

Solve for y.

Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.

Exercise $\PageIndex{20}$

Solve the system by substitution. $\left\{\begin{array}{l}{2 x-3 y=12} \\ {-12 y+8 x=48}\end{array}\right.$

Answer

infinitely many solutions

Exercise $\PageIndex{21}$

Solve the system by substitution. $\left\{\begin{array}{l}{5 x+2 y=12} \\ {-4 y-10 x=-24}\end{array}\right.$

Answer

infinitely many solutions

Look back at the equations in Exercise $\PageIndex{22}$. Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Exercise $\PageIndex{22}$

Solve the system by substitution. $\left\{\begin{array}{l}{5 x-2 y=-10} \\ {y=\frac{5}{2} x}\end{array}\right.$

Answer

The second equation is already solved for y, so we can substitute for y in the first equation.

Substitute x for y in the first equation.

Replace the y with $\frac{5}{2}x$.

Solve for x.

Since 0 = −10 is a false statement the equations are inconsistent. The graphs of the two equation would be parallel lines. The system has no solutions.

Exercise $\PageIndex{23}$

Solve the system by substitution. $\left\{\begin{array}{l}{3 x+2 y=9} \\ {y=-\frac{3}{2} x+1}\end{array}\right.$

Answer

no solution

Exercise $\PageIndex{24}$

Solve the system by substitution. $\left\{\begin{array}{l}{5 x-3 y=2} \\ {y=\frac{5}{3} x-4}\end{array}\right.$

Answer

no solution

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

HOW TO USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose variables to represent those quantities.
Translate into a system of equations.
Solve the system of equations using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Exercise $\PageIndex{25}$

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

AnswerStep 1. Read the problem. Step 2. Identify what we are looking for.We are looking for two numbers.Step 3. Name what we are looking for.Let n= the first number
Let m= the second numberStep 4. Translate into a system of equations.The sum of two numbers is zero.

One number is nine less than the other.

The system is:

Step 5. Solve the system of
equations. We will use substitution
since the second equation is solved
for n. Substitute m − 9 for n in the first equation.

Solve for m.

Substitute $m=\frac{9}{2}$ into the second equation
and then solve for n.

Step 6. Check the answer in the problem.Do these numbers make sense in
the problem? We will leave this to you!Step 7. Answer the question.The numbers are $\frac{9}{2}$ and $-\frac{9}{2}$.

Exercise $\PageIndex{26}$

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Answer

The numbers are 3 and 7.

Exercise $\PageIndex{27}$

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

Answer

The numbers are 2 and −8.

In the Exercise $\PageIndex{28}$, we’ll use the formula for the perimeter of a rectangle, P = 2L + 2W.

Exercise $\PageIndex{28}$

Add exercises text here.

AnswerStep 1. Read the problem.

Step 2. Identify what you are looking for.We are looking for the length and width.Step 3. Name what we are looking for.Let L= the length
W= the widthStep 4. Translate into a system of equations.The perimeter of a rectangle is 88. 2L + 2W = P

The length is five more than twice the width.

The system is:

Step 5. Solve the system of equations.
We will use substitution since the second
equation is solved for L.

Substitute 2W + 5 for L in the first equation.

Solve for W.

Substitute W = 13 into the second
equation and then solve for L.

Step 6. Check the answer in the problem.Does a rectangle with length 31 and width
13 have perimeter 88? Yes.Step 7. Answer the equation.The length is 31 and the width is 13.

Exercise $\PageIndex{29}$

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

Answer

The length is 12 and the width is 8.

Exercise $\PageIndex{30}$

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

Answer

The length is 23 and the width is 6.

For Exercise $\PageIndex{31}$ we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Exercise $\PageIndex{31}$

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

Answer

We will draw and label a figure.

Step 1. Read the problem.

Step 2. Identify what you are looking for.We are looking for the measures of the angles.Step 3. Name what we are looking for.Let a= the measure of the 1st angle
b= the measure of the 2nd angleStep 4. Translate into a system of equations.The measure of one of the small angles
of a right triangle is ten more than three
times the measure of the other small angle.

The sum of the measures of the angles of
a triangle is 180.

The system is:

Step 5. Solve the system of equations.
We will use substitution since the first
equation is solved for a.

Substitute 3b + 10 for a in the
second equation.

Solve for b.

Substitute b = 20 into the first
equation and then solve for a.

Step 6. Check the answer in the problem.We will leave this to you!Step 7. Answer the question.The measures of the small angles are
20 and 70.

Exercise $\PageIndex{32}$

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Answer

The measure of the angles are 22 degrees and 68 degrees.

Exercise $\PageIndex{33}$

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Answer

The measure of the angles are 36 degrees and 54 degrees.

Exercise $\PageIndex{34}$

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

AnswerStep 1. Read the problem. Step 2. Identify what you are looking for.We are looking for the number of training sessions
that would make the pay equal.Step 3. Name what we are looking for.Let s= Heather’s salary.
n= the number of training sessionsStep 4. Translate into a system of equations.Option A would pay her $25,000 plus $15
for each training session.

Option B would pay her $10,000 + $40
for each training session

The system is:

Step 5. Solve the system of equations.
We will use substitution.

Substitute 25,000 + 15n for s in the second equation.

Solve for n.

Step 6. Check the answer.Are 600 training sessions a year reasonable?
Are the two options equal when n = 600?Step 7. Answer the question.The salary options would be equal for 600 training sessions.

Exercise $\PageIndex{35}$

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

Answer

There would need to be 160 policies sold to make the total pay the same.

Exercise $\PageIndex{36}$

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Answer

Kenneth would need to sell 1,000 suits.

Note

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

Instructional Video-Solve Linear Systems by Substitution
Instructional Video-Solve by Substitution

Key Concepts

Solve a system of equations by substitution
1. Solve one of the equations for either variable.
2. Substitute the expression from Step 1 into the other equation.
3. Solve the resulting equation.
4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
5. Write the solution as an ordered pair.
6. Check that the ordered pair is a solution to both original equations.

This page titled 4.2: Solve Systems of Equations by Substitution is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.

How do you solve by substitution step by step?

Steps to Solving by Substitution:.

Step One→ Solve one equation for either x or y..

Step Two→ Substitute the expression from step one into the 2nd equation..

Step Three→ Solve the second equation for the given variable..

Step Four→ Plug you solution back into the first equation..

Step Five→ Write your solution as a point..

What are the three steps when solving system of equations by substitution?

We can summarize the method of substitution in three steps: Solve any one equation for one of its variables. Put/Substitute the found value into the other equation and solve it. Put/Substitute the solution for one variable into the original equation to find the other variable.

How do you solve two equations with substitution?

To solve systems using substitution, follow this procedure:.

Select one equation and solve it for one of its variables..

In the other equation, substitute for the variable just solved..

Solve the new equation..

Substitute the value found into any equation involving both variables and solve for the other variable..

Which system of equations would be best solved using substitution?

Substitution is best used when one of the equations is in terms of one of the variables such as y=2x+4, but the equations can always be manipulated. The first step in this method is to solve one of the equations for one variable.

How to solve system of equations using substitution

Learning Objectives

Note

Solve a System of Equations by Substitution

Exercise \(\PageIndex{1}\): How to Solve a System of Equations by Substitution

Exercise \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)

SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION.

Exercise \(\PageIndex{4}\)

Exercise \(\PageIndex{5}\)

Exercise \(\PageIndex{6}\)

Exercise \(\PageIndex{7}\)

Exercise \(\PageIndex{8}\)

Exercise \(\PageIndex{9}\)

Exercise \(\PageIndex{10}\)

Exercise \(\PageIndex{11}\)

Exercise \(\PageIndex{12}\)

Exercise \(\PageIndex{13}\)

Exercise \(\PageIndex{14}\)

Exercise \(\PageIndex{15}\)

Exercise \(\PageIndex{16}\)

Exercise \(\PageIndex{17}\)

Exercise \(\PageIndex{18}\)

Exercise \(\PageIndex{19}\)

Exercise \(\PageIndex{20}\)

Exercise \(\PageIndex{21}\)

Exercise \(\PageIndex{22}\)

Exercise \(\PageIndex{23}\)

Exercise \(\PageIndex{24}\)

Solve Applications of Systems of Equations by Substitution

HOW TO USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.

Exercise \(\PageIndex{25}\)

Exercise \(\PageIndex{26}\)

Exercise \(\PageIndex{27}\)

Exercise \(\PageIndex{28}\)

Exercise \(\PageIndex{29}\)

Exercise \(\PageIndex{30}\)

Exercise \(\PageIndex{31}\)

Exercise \(\PageIndex{32}\)

Exercise \(\PageIndex{33}\)

Exercise \(\PageIndex{34}\)

Exercise \(\PageIndex{35}\)

Exercise \(\PageIndex{36}\)

Note

Key Concepts

How do you solve by substitution step by step?

What are the three steps when solving system of equations by substitution?

How do you solve two equations with substitution?

Which system of equations would be best solved using substitution?

Related Posts

Advertising

LATEST NEWS

Advertising

Populer

Advertising

About

Legal

Help

Social