# 5.3 Solve systems of equations by elimination  (Page 3/6)

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

$\left(1,3\right)$

Solve the system by elimination. $\left\{\begin{array}{c}7x+8y=4\hfill \\ 3x-5y=27\hfill \end{array}$

$\left(4,-3\right)$

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.

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

## Solution

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. $\begin{array}{cccc}\begin{array}{ccc}\hfill x+\frac{1}{2}y& =\hfill & 6\hfill \\ \hfill 3+\frac{1}{2}\left(6\right)& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 3+6& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 6& =\hfill & 6\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \\ \\ \\ \\ \end{array}& & & \begin{array}{ccc}\hfill \frac{3}{2}x+\frac{2}{3}y& =\hfill & \frac{17}{2}\hfill \\ \hfill \frac{3}{2}\left(3\right)+\frac{2}{3}\left(6\right)& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+4& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+\frac{8}{2}& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{17}{2}& =\hfill & \frac{17}{2}\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$ The solution is (3, 6).

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

$\left(6,2\right)$

Solve the system by elimination. $\left\{\begin{array}{c}x+\frac{3}{5}y=-\frac{1}{5}\hfill \\ -\frac{1}{2}x-\frac{2}{3}y=\frac{5}{6}\hfill \end{array}$

$\left(1,-2\right)$

In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

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

## Solution

$\begin{array}{ccc}& & \phantom{\rule{0.8em}{0ex}}\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill y& =\hfill & 3-\frac{3}{4}x\hfill \end{array}\hfill \\ \\ \\ \text{Write the second equation in standard form.}\hfill & & \phantom{\rule{0.8em}{0ex}}\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill \frac{3}{4}x+y& =\hfill & 3\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{Clear the fractions by multiplying the}\hfill \\ \text{second equation by 4.}\hfill \end{array}\hfill & & \left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 4\left(\frac{3}{4}x+y\right)& =\hfill & 4\left(3\right)\hfill \end{array}\hfill \\ \\ \\ \text{Simplify.}\hfill & & \phantom{\rule{1em}{0ex}}\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 3x+4y& =\hfill & 12\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{To eliminate a variable, we multiply the}\hfill \\ \text{second equation by −1.}\hfill \end{array}\hfill & & \begin{array}{c}\phantom{\rule{0.2em}{0ex}}\underset{\text{________________}}{\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill -3x-4y& =\hfill & -12\hfill \end{array}}\hfill \\ \hfill 0=0\hfill \end{array}\hfill \\ \text{Simplify and add.}\hfill & \end{array}$

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. $\left\{\begin{array}{c}5x-3y=15\hfill \\ y=-5+\frac{5}{3}x\hfill \end{array}$

infinitely many solutions

Solve the system by elimination. $\left\{\begin{array}{c}x+2y=6\hfill \\ y=-\frac{1}{2}x+3\hfill \end{array}$

infinitely many solutions

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

## Solution

$\begin{array}{cc}\text{The equations are in standard form.}\hfill & \phantom{\rule{0.1em}{0ex}}\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & 10\hfill \\ \hfill 2x-5y& =\hfill & -5\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{Multiply the second equation by 3 to}\hfill \\ \text{eliminate a variable.}\hfill \end{array}\hfill & \phantom{\rule{0.1em}{0ex}}\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & 10\hfill \\ \hfill 3\left(2x-5y\right)& =\hfill & 3\left(-5\right)\hfill \end{array}\hfill \\ \\ \\ \text{Simplify and add.}\hfill & \begin{array}{c}\underset{\text{__________________}}{\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & \phantom{\rule{0.5em}{0ex}}10\hfill \\ \hfill 6x-15y& =\hfill & -15\hfill \end{array}}\\ \hfill 0\ne -5\hfill \end{array}\hfill \end{array}$

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. $\left\{\begin{array}{c}-3x+2y=8\hfill \\ 9x-6y=13\hfill \end{array}$

no solution

Solve the system by elimination. $\left\{\begin{array}{c}7x-3y=-2\hfill \\ -14x+6y=8\hfill \end{array}$

no solution

## 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.

#### Questions & Answers

Larry and Tom were standing next to each other in the backyard when Tom challenged Larry to guess how tall he was. Larry knew his own height is 6.5 feet and when they measured their shadows, Larry’s shadow was 8 feet and Tom’s was 7.75 feet long. What is Tom’s height?
6.25
Ciid
6.25
Big
Wayne is hanging a string of lights 57 feet long around the three sides of his patio, which is adjacent to his house. the length of his patio, the side along the house, is 5 feet longer than twice it's width. Find the length and width of the patio.
Ciid
tyler
(sin=opp/adj) (tan= opp/adj) cos=hyp/adj dont quote me on it look it up
tyler
(sin=opp/adj) (tan= opp/adj) cos=hyp/adj dont quote me on it look it up
tyler
(sin=opp/adj) (tan= opp/adj) cos=hyp/adj dont quote me on it look it up
tyler
SOH = Sine is Opposite over Hypotenuse. CAH= Cosine is Adjacent over Hypotenuse. TOA = Tangent is Opposite over Adjacent.
tyler
H=57 and O=285 figure out what the adjacent?
tyler
Amara currently sells televisions for company A at a salary of $17,000 plus a$100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a$20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?
what is the quantity and price of the televisions for both options?
karl
Amara has to sell 120 televisions to make 29,000 of the salary of company B. 120 * TV 100= commission 12,000+ 17,000 = of company a salary 29,000 17000+
Ciid
Amara has to sell 120 televisions to make 29,000 of the salary of company B. 120 * TV 100= commission 12,000+ 17,000 = of company a salary 29,000
Ciid
I'm mathematics teacher from highly recognized university.
here a question professor How many soldiers are there in a group of 27 sailors and soldiers if there are four fifths as many sailors as soldiers? can you write out the college you went to with the name of the school you teach at and let me know the answer I've got it to be honest with you
tyler
is anyone else having issues with the links not doing anything?
Yes
Val
chapter 1 foundations 1.2 exercises variables and algebraic symbols
June needs 45 gallons of punch for a party and has 2 different coolers to carry it in. The bigger cooler is 5 times as large as the smaller cooler. How many gallons can each cooler hold? Enter the answers in decimal form.
Joseph would like to make 12 pounds of a coffee blend at a cost of $6.25 per pound. He blends Ground Chicory at$4.40 a pound with Jamaican Blue Mountain at $8.84 per pound. How much of each type of coffee should he use? Samer 4x6.25=$25 coffee blend 4×4.40= $17.60 ground chicory 4x8.84= 35.36 blue mountain. In total they will spend for 12 pounds$77.96 they will spend in total
tyler
DaMarcus and Fabian live 23 miles apart and play soccer at a park between their homes. DaMarcus rode his bike for three-quarters of an hour and Fabian rode his bike for half an hour to get to the park. Fabian’s speed was six miles per hour faster than DaMarcus’ speed. Find the speed of both soccer players.
i need help how to do this is confusing
what kind of math is it?
Danteii
help me to understand
huh, what is the algebra problem
Daniel
How many soldiers are there in a group of 27 sailors and soldiers if there are four fifths many sailors as soldiers?
tyler
What is the domain and range of heaviside
What is the domain and range of Heaviside and signum
Christopher
25-35
Fazal
The hypotenuse of a right triangle is 10cm long. One of the triangle’s legs is three times the length of the other leg. Find the lengths of the three sides of the triangle.
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A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of the 50% solution and how many liters of the 90% solution will be used?
June needs 45 gallons of punch for a party and has 2 different coolers to carry it in. The bigger cooler is 5 times as large as the smaller cooler. How many gallons can each cooler hold? By   By  By Lakeima Roberts    By Qqq Qqq