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Solve Systems of Equations by Elimination

Solve a system of equations by elimination

Solve a System of Equations by Elimination

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

\[\begin{array}{llll}\text{if} & a & = & b \\ \text{and} & c & = & d \\ \text{then} & a+c & = & b+d\end{array}\]

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

\[\begin{array}{l}3x+y=5 \\ \underset{\text{_________}}{2x-y=0} \\ 5x\ =5\end{array}\]

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

\[\{\begin{array}{l}x+4y=2 \\ 2x+5y=-2\end{array}\]

This time we don’t see a variable that can be immediately eliminated if we add the equations.

\[\{\begin{array}{l}4x-3y=10 \\ 3x+5y=-7\end{array}\]\[\{\begin{array}{l}\ 12x-9y=30 \\ -12x-20y=28\end{array}\]\[\begin{array}{l}\{\begin{array}{l}\ 12x-9y=30 \\ \underset{\text{_____________}}{-12x-20y=28}\end{array} \\ -29y=58\end{array}\]

How to Solve a System of Equations by Elimination

Try it.

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

Solution

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

Solve Applications of Systems of Equations by Elimination

Some applications problems translate directly into equations in standard form, so we will use the elimination method to solve them. As before, we use our Problem Solving Strategy to help us stay focused and organized.

Example

Try it.

The sum of two numbers is 39. Their difference is 9. Find the numbers.

Solution

Step 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. Choose a variable to represent that quantity.	Let $n=$ the first number. $m=$ the second number.
Step 4. Translate into a system of equations. The system is:	The sum of two numbers is 39. $n+m=39$ Their difference is 9. $\begin{array}{l}n-m=9 \\ \{\begin{array}{l}n+m=39 \\ n-m=9\end{array}\end{array}$
Step 5. Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form and the coefficients of $m$ are opposites. Add. Solve for $n$. Substitute $n=24$ into one of the original equations and solve for $m$.	$\begin{array}{l}\underset{\text{____________}}{\{\begin{array}{l}n+m=39 \\ n-m=9\end{array}} \\ 2n\ =48 \\ \\ n=24 \\ n+m=39 \\ 24+m=39 \\ m=15\end{array}$
Step 6. Check the answer.	Since $24+15=39$ and $24-15=9$, the answers check.
Step 7. Answer the question.	The numbers are 24 and 15.

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

Choose the Most Convenient Method to Solve a System of Linear Equations

When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example

Try it.

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ $\{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}$ ⓑ $\{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}$

Solution

ⓐ $\begin{array}{lll} & & \{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}\end{array}$
Since both equations are in standard form, using elimination will be most convenient.
ⓑ $\begin{array}{lll} & & \{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}\end{array}$

>Since one equation is already solved for y, using substitution will be most convenient.

Key Concepts

To Solve a System of Equations by Elimination
1. Write both equations in standard form. If any coefficients are fractions, clear them.
2. Make the coefficients of one variable opposites.
  - Decide which variable you will eliminate.
  - Multiply one or both equations so that the coefficients of that variable are opposites.
3. Add the equations resulting from Step 2 to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
6. Write the solution as an ordered pair.
7. Check that the ordered pair is a solution to both original equations.

Solve Systems of Equations by Elimination

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

Try it.

$\{\begin{array}{l}5x+2y=2 \\ -3x-y=0\end{array}$

Try it.

$\{\begin{array}{l}-3x+y=-9 \\ x-2y=-12\end{array}$

Solution

(6, 9)

Try it.

$\{\begin{array}{l}6x-5y=-1 \\ 2x+y=13\end{array}$

Try it.

$\{\begin{array}{l}3x-y=-7 \\ 4x+2y=-6\end{array}$

Solution

$(-2,1)$

Try it.

$\{\begin{array}{l}x+y=-1 \\ x-y=-5\end{array}$

Try it.

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

Solution

$(-7,-1)$

Try it.

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

Try it.

$\{\begin{array}{l}-7x+6y=-10 \\ x-6y=22\end{array}$

Solution

$(-2,-4)$

Try it.

$\{\begin{array}{l}3x+2y=-3 \\ -x-2y=-19\end{array}$

Try it.

$\{\begin{array}{l}5x+2y=1 \\ -5x-4y=-7\end{array}$

Solution

$(-1,3)$

Try it.

$\{\begin{array}{l}6x+4y=-4 \\ -6x-5y=8\end{array}$

Try it.

$\{\begin{array}{l}3x-4y=-11 \\ x-2y=-5\end{array}$

Solution

$(-1,2)$

Try it.

$\{\begin{array}{l}5x-7y=29 \\ x+3y=-3\end{array}$

Try it.

$\{\begin{array}{l}6x-5y=-75 \\ -x-2y=-13\end{array}$

Solution

$(-5,9)$

Try it.

$\{\begin{array}{l}-x+4y=8 \\ 3x+5y=10\end{array}$

Try it.

$\{\begin{array}{l}2x-5y=7 \\ 3x-y=17\end{array}$

Solution

(6, 1)

Try it.

$\{\begin{array}{l}5x-3y=-1 \\ 2x-y=2\end{array}$

Try it.

$\{\begin{array}{l}7x+y=-4 \\ 13x+3y=4\end{array}$

Solution

$(-2,10)$

Try it.

$\{\begin{array}{l}-3x+5y=-13 \\ 2x+y=-26\end{array}$

Try it.

$\{\begin{array}{l}3x-5y=-9 \\ 5x+2y=16\end{array}$

Solution

(2, 3)

Try it.

$\{\begin{array}{l}4x-3y=3 \\ 2x+5y=-31\end{array}$

Try it.

$\{\begin{array}{l}4x+7y=14 \\ -2x+3y=32\end{array}$

Solution

$(-7,6)$

Try it.

$\{\begin{array}{l}5x+2y=21 \\ 7x-4y=9\end{array}$

Try it.

$\{\begin{array}{l}3x+8y=-3 \\ 2x+5y=-3\end{array}$

Solution

$(-9,3)$

Try it.

$\{\begin{array}{l}11x+9y=-5 \\ 7x+5y=-1\end{array}$

Try it.

$\{\begin{array}{l}3x+8y=67 \\ 5x+3y=60\end{array}$

Solution

(9, 5)

Try it.

$\{\begin{array}{l}2x+9y=-4 \\ 3x+13y=-7\end{array}$

Try it.

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

Solution

$(-3,2)$

Try it.

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

Try it.

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

Solution

$(-2,3)$

Try it.

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

Try it.

$\{\begin{array}{l}2x+y=3 \\ 6x+3y=9\end{array}$

Solution

infinitely many solutions

Try it.

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

Try it.

$\{\begin{array}{l}-3x-y=8 \\ 6x+2y=-16\end{array}$

Solution

infinitely many solutions

Try it.

$\{\begin{array}{l}4x+3y=2 \\ 20x+15y=10\end{array}$

Try it.

$\{\begin{array}{l}3x+2y=6 \\ -6x-4y=-12\end{array}$

Solution

infinitely many solutions

Try it.

$\{\begin{array}{l}5x-8y=12 \\ 10x-16y=20\end{array}$

Try it.

$\{\begin{array}{l}-11x+12y=60 \\ -22x+24y=90\end{array}$

Solution

inconsistent, no solution

Try it.

$\{\begin{array}{l}7x-9y=16 \\ -21x+27y=-24\end{array}$

Try it.

$\{\begin{array}{l}5x-3y=15 \\ y=\frac{5}{3}x-2\end{array}$

Solution

inconsistent, no solution

Try it.

$\{\begin{array}{l}2x+4y=7 \\ y=-\frac{1}{2}x-4\end{array}$

Solve Applications of Systems of Equations by Elimination

In the following exercises, translate to a system of equations and solve.

Try it.

The sum of two numbers is 65. Their difference is 25. Find the numbers.

Solution

The numbers are 20 and 45.

Try it.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

Try it.

The sum of two numbers is −27. Their difference is −59. Find the numbers.

Solution

The numbers are 16 and −43.

Try it.

The sum of two numbers is −45. Their difference is −89. Find the numbers.

Try it.

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Solution

A shirt costs $16 and a sweater costs $33.

Try it.

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

Try it.

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

Solution

There are 860 mg in a hot dog. There are 1,000 mg in a cup of cottage cheese.

Try it.

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

Choose the Most Convenient Method to Solve a System of Linear Equations

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

Try it.

ⓐ $\{\begin{array}{l}8x-15y=-32 \\ 6x+3y=-5\end{array}$ ⓑ $\{\begin{array}{l}x=4y-3 \\ 4x-2y=-6\end{array}$

Solution

ⓐ elimination ⓑ substitution

Try it.

ⓐ $\{\begin{array}{l}y=7x-5 \\ 3x-2y=16\end{array}$ ⓑ $\{\begin{array}{l}12x-5y=-42 \\ 3x+7y=-15\end{array}$

Try it.

ⓐ $\{\begin{array}{l}y=4x+9 \\ 5x-2y=-21\end{array}$ ⓑ $\{\begin{array}{l}9x-4y=24 \\ 3x+5y=-14\end{array}$

Solution

ⓐ substitution ⓑ elimination

Try it.

ⓐ $\{\begin{array}{l}14x-15y=-30 \\ 7x+2y=10\end{array}$ ⓑ $\{\begin{array}{l}x=9y-11 \\ 2x-7y=-27\end{array}$

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.

Simplify $-5(6-3a)$.
If you missed this problem, review .

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$-30+15a$
Solve the equation $\frac{1}{3}x+\frac{5}{8}=\frac{31}{24}$.
If you missed this problem, review .

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$x=2$
Solve the system by elimination. $\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}$
Solve the system by elimination. $\{\begin{array}{l}3x+y=5 \\ 2x-3y=7\end{array}$

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$(2,-1)$
Solve the system by elimination. $\{\begin{array}{l}4x+y=-5 \\ -2x-2y=-2\end{array}$

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$(-2,3)$

Solve the system by elimination. $\{\begin{array}{l}x+y=10 \\ x-y=12\end{array}$

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Both equations are in standard form.
The coefficients of y are already opposites.
Add the two equations to eliminate y. The resulting equation has only 1 variable, x.
Solve for x, the remaining variable. Substitute x = 11 into one of the original equations.

Solve for the other variable, y.
Write the solution as an ordered pair.	The ordered pair is (11, −1).
Check that the ordered pair is a solution to both original equations. $\begin{array}{llllllllllllllll}\begin{array}{lll}x+y & = & 10 \\ 11+(-1) & \overset{?}{=} & 10 \\ 10 & = & 10\ ✓\end{array} & & & \begin{array}{lll}x-y & = & 12 \\ 11-(-1) & \overset{?}{=} & 12 \\ 12 & = & 12\ ✓\end{array}\end{array}$
	The solution is (11, −1).

Solve the system by elimination. $\{\begin{array}{l}2x+y=5 \\ x-y=4\end{array}$

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

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$(-2,5)$

Solve the system by elimination. $\{\begin{array}{l}3x-2y=-2 \\ 5x-6y=10\end{array}$

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Both equations are in standard form.
None of the coefficients are opposites.
We can make the coefficients of y opposites by multiplying the first equation by −3.
Simplify.
Add the two equations to eliminate y.
Solve for the remaining variable, x. Substitute x = −4 into one of the original equations.

Solve for y.
Write the solution as an ordered pair.	The ordered pair is (−4, −5).
Check that the ordered pair is a solution to both original equations. $\begin{array}{llllllllllllllllllll}\begin{array}{lll}3x-2y & = & -2 \\ 3(-4)-2(-5) & \overset{?}{=} & -2 \\ -12+10 & \overset{?}{=} & -2 \\ -2y & = & -2\ ✓\end{array} & & & \begin{array}{lll}5x-6y & = & 10 \\ 3(-4)-6(-5) & \overset{?}{=} & 10 \\ -20+30 & \overset{?}{=} & 10 \\ 10 & = & 10\ ✓\end{array}\end{array}$
	The solution is (−4, −5).

Solve the system by elimination. $\{\begin{array}{l}4x-3y=1 \\ 5x-9y=-4\end{array}$

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$(1,1)$
Solve the system by elimination. $\{\begin{array}{l}3x+2y=2 \\ 6x+5y=8\end{array}$

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$(-2,4)$

Solve the system by elimination. $\{\begin{array}{l}4x-3y=9 \\ 7x+2y=-6\end{array}$

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In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.


Both equations are in standard form. To get opposite coefficients of y, we will multiply the first equation by 2 and the second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x. Substitute x = 0 into one of the original equations.

Solve for y.

Write the solution as an ordered pair.	The ordered pair is (0, −3).
Check that the ordered pair is a solution to both original equations. $\begin{array}{llllllllllllllll}\begin{array}{lll}4x-3y & = & 9 \\ 4(0)-3(-3) & \overset{?}{=} & 9 \\ 9 & = & 9\ ✓\end{array} & & & \begin{array}{lll}7x+2y & = & -6 \\ 7(0)+2(-3) & \overset{?}{=} & -6 \\ -6 & = & -6\ ✓\end{array}\end{array}$
	The solution is (0, −3).

What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?

Solve the system by elimination. $\{\begin{array}{l}3x-4y=-9 \\ 5x+3y=14\end{array}$

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$(1,3)$
Solve the system by elimination. $\{\begin{array}{l}7x+8y=4 \\ 3x-5y=27\end{array}$

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$(4,-3)$

Solve the system by elimination. $\{\begin{array}{l}x+\frac{1}{2}y=6 \\ \frac{3}{2}x+\frac{2}{3}y=\frac{17}{2}\end{array}$

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In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.


To clear the fractions, multiply each equation by its LCD.
Simplify.
Now we are ready to eliminate one of the variables. Notice that both equations are in standard form.
We can eliminate y multiplying the top equation by −4.
Simplify and add. Substitute x = 3 into one of the original equations.
Solve for y.


Write the solution as an ordered pair.	The ordered pair is (3, 6).
Check that the ordered pair is a solution to both original equations. $\begin{array}{llllllllllllllllllllll}\begin{array}{lll}x+\frac{1}{2}y & = & 6 \\ 3+\frac{1}{2}(6) & \overset{?}{=} & 6 \\ 3+3 & \overset{?}{=} & 6 \\ 6 & = & 6\ ✓ \\ \\ \\ \\ \\ \end{array} & & & \begin{array}{lll}\frac{3}{2}x+\frac{2}{3}y & = & \frac{17}{2} \\ \frac{3}{2}(3)+\frac{2}{3}(6) & \overset{?}{=} & \frac{17}{2} \\ \frac{9}{2}+4 & \overset{?}{=} & \frac{17}{2} \\ \frac{9}{2}+\frac{8}{2} & \overset{?}{=} & \frac{17}{2} \\ \frac{17}{2} & = & \frac{17}{2}\ ✓\end{array}\end{array}$
	The solution is (3, 6).

Solve the system by elimination. $\{\begin{array}{l}\frac{1}{3}x-\frac{1}{2}y=1 \\ \frac{3}{4}x-y=\frac{5}{2}\end{array}$

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$(6,2)$
Solve the system by elimination. $\{\begin{array}{l}x+\frac{3}{5}y=-\frac{1}{5} \\ -\frac{1}{2}x-\frac{2}{3}y=\frac{5}{6}\end{array}$

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$(1,-2)$

Solve the system by elimination. $\{\begin{array}{l}3x+4y=12 \\ y=3-\frac{3}{4}x\end{array}$

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	$\{\begin{array}{l}3x+4y=12 \\ y=3-\frac{3}{4}x\end{array}$
Write the second equation in standard form.	$\{\begin{array}{lll}3x+4y & = & 12 \\ \frac{3}{4}x+y & = & 3\end{array}$
Clear the fractions by multiplying the second equation by 4.	$\{\begin{array}{lll}3x+4y & = & 12 \\ 4(\frac{3}{4}x+y) & = & 4(3)\end{array}$
Simplify.	$\{\begin{array}{lll}3x+4y & = & 12 \\ 3x+4y & = & 12\end{array}$
To eliminate a variable, we multiply the second equation by $-1$. Simplify and add.	$\begin{array}{lllll}\ \underset{\text{________________}}{\{\begin{array}{lll}3x+4y & = & 12 \\ -3x-4y & = & -12\end{array}} \\ 0=0\end{array}$

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Solve the system by elimination. $\{\begin{array}{l}5x-3y=15 \\ y=-5+\frac{5}{3}x\end{array}$

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infinitely many solutions
Solve the system by elimination. $\{\begin{array}{l}x+2y=6 \\ y=-\frac{1}{2}x+3\end{array}$

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infinitely many solutions

Solve the system by elimination. $\{\begin{array}{l}-6x+15y=10 \\ 2x-5y=-5\end{array}$

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The equations are in standard form.	$\{\begin{array}{lll}-6x+15y & = & 10 \\ 2x-5y & = & -5\end{array}$
Multiply the second equation by 3 to eliminate a variable.	$\{\begin{array}{lll}-6x+15y & = & 10 \\ 3(2x-5y) & = & 3(-5)\end{array}$
Simplify and add.	$\begin{array}{lllll}\underset{\text{__________________}}{\{\begin{array}{lll}-6x+15y & = & \ 10 \\ 6x-15y & = & -15\end{array}} \\ 0\ne -5\end{array}$

This statement is false. The equations are inconsistent and so their graphs would be parallel lines.

The system does not have a solution.

Solve the system by elimination. $\{\begin{array}{l}-3x+2y=8 \\ 9x-6y=13\end{array}$

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no solution
Solve the system by elimination. $\{\begin{array}{l}7x-3y=-2 \\ -14x+6y=8\end{array}$

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no solution

The sum of two numbers is 39. Their difference is 9. Find the numbers.

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Step 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. Choose a variable to represent that quantity.	Let $n=$ the first number. $m=$ the second number.
Step 4. Translate into a system of equations. The system is:	The sum of two numbers is 39. $n+m=39$ Their difference is 9. $\begin{array}{l}n-m=9 \\ \{\begin{array}{l}n+m=39 \\ n-m=9\end{array}\end{array}$
Step 5. Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form and the coefficients of $m$ are opposites. Add. Solve for $n$. Substitute $n=24$ into one of the original equations and solve for $m$.	$\begin{array}{l}\underset{\text{____________}}{\{\begin{array}{l}n+m=39 \\ n-m=9\end{array}} \\ 2n\ =48 \\ \\ n=24 \\ n+m=39 \\ 24+m=39 \\ m=15\end{array}$
Step 6. Check the answer.	Since $24+15=39$ and $24-15=9$, the answers check.
Step 7. Answer the question.	The numbers are 24 and 15.

The sum of two numbers is 42. Their difference is 8. Find the numbers.

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The numbers are 25 and 17.
The sum of two numbers is −15. Their difference is −35. Find the numbers.

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The numbers are −25 and 10.

Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?

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Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for the number of calories in one order of medium fries and in one small soda.
Step 3. Name what we are looking for.	Let f = the number of calories in 1 order of medium fries. s = the number of calories in 1 small soda.
Step 4. Translate into a system of equations:	one medium fries and two small sodas had a total of 620 calories

	two medium fries and one small soda had a total of 820 calories.

Our system is:
Step 5. Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form. To get opposite coefficients of f, multiply the top equation by −2.
Simplify and add.
Solve for s.
Substitute s = 140 into one of the original equations and then solve for f.



Step 6. Check the answer.	Verify that these numbers make sense in the problem and that they are solutions to both equations. We leave this to you!
Step 7. Answer the question.	The small soda has 140 calories and the fries have 340 calories.

Malik stops at the grocery store to buy a bag of diapers and 2 cans of formula. He spends a total of $37. The next week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a bag of diapers cost? How much is one can of formula?

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The bag of diapers costs $11 and the can of formula costs $13.
To get her daily intake of fruit for the day, Sasha eats a banana and 8 strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats two bananas and 5 strawberries for a total of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?

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There are 105 calories in a banana and 5 calories in a strawberry.
For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ $\{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}$ ⓑ $\{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}$
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1. ⓐ $\begin{array}{lll} & & \{\begin{array}{l}3x+8y=40 \\ 7x-4y=-32\end{array}\end{array}$
  Since both equations are in standard form, using elimination will be most convenient.
2. ⓑ $\begin{array}{lll} & & \{\begin{array}{l}5x+6y=12 \\ y=\frac{2}{3}x-1\end{array}\end{array}$
>Since one equation is already solved for y, using substitution will be most convenient.
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ $\{\begin{array}{l}4x-5y=-32 \\ 3x+2y=-1\end{array}$ ⓑ $\{\begin{array}{l}x=2y-1 \\ 3x-5y=-7\end{array}$

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ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for $x$, using substitution will be most convenient.
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ $\{\begin{array}{l}y=2x-1 \\ 3x-4y=-6\end{array}$ ⓑ $\{\begin{array}{l}6x-2y=12 \\ 3x+7y=-13\end{array}$

Разкрийте отговора

ⓐ Since one equation is already solved for $y$, using substitution will be most convenient; ⓑ Since both equations are in standard form, using elimination will be most convenient.
$\{\begin{array}{l}5x+2y=2 \\ -3x-y=0\end{array}$
$\{\begin{array}{l}-3x+y=-9 \\ x-2y=-12\end{array}$

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(6, 9)
$\{\begin{array}{l}6x-5y=-1 \\ 2x+y=13\end{array}$
$\{\begin{array}{l}3x-y=-7 \\ 4x+2y=-6\end{array}$

Разкрийте отговора

$(-2,1)$
$\{\begin{array}{l}x+y=-1 \\ x-y=-5\end{array}$
$\{\begin{array}{l}x+y=-8 \\ x-y=-6\end{array}$

Разкрийте отговора

$(-7,-1)$
$\{\begin{array}{l}3x-2y=1 \\ -x+2y=9\end{array}$
$\{\begin{array}{l}-7x+6y=-10 \\ x-6y=22\end{array}$

Разкрийте отговора

$(-2,-4)$

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 Elimination

Solve a system of equations by elimination
Solve applications of systems of equations by elimination
Choose the most convenient method to solve a system of linear equations
Write both equations in standard form. If any coefficients are fractions, clear them.
Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that variable are opposites.
Add the equations resulting from Step 2 to eliminate one variable.

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.

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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.

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	\(\{\begin{array}{l}3x+4y=12 \\ y=3-\frac{3}{4}x\end{array}\)
Write the second equation in standard form.	\(\{\begin{array}{lll}3x+4y & = & 12 \\ \frac{3}{4}x+y & = & 3\end{array}\)
Clear the fractions by multiplying the second equation by 4.	\(\{\begin{array}{lll}3x+4y & = & 12 \\ 4(\frac{3}{4}x+y) & = & 4(3)\end{array}\)
Simplify.	\(\{\begin{array}{lll}3x+4y & = & 12 \\ 3x+4y & = & 12\end{array}\)
To eliminate a variable, we multiply the second equation by \(-1\). Simplify and add.	\(\begin{array}{lllll}\ \underset{\text{________________}}{\{\begin{array}{lll}3x+4y & = & 12 \\ -3x-4y & = & -12\end{array}} \\ 0=0\end{array}\)

The equations are in standard form.	\(\{\begin{array}{lll}-6x+15y & = & 10 \\ 2x-5y & = & -5\end{array}\)
Multiply the second equation by 3 to eliminate a variable.	\(\{\begin{array}{lll}-6x+15y & = & 10 \\ 3(2x-5y) & = & 3(-5)\end{array}\)
Simplify and add.	\(\begin{array}{lllll}\underset{\text{__________________}}{\{\begin{array}{lll}-6x+15y & = & \ 10 \\ 6x-15y & = & -15\end{array}} \\ 0\ne -5\end{array}\)