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

 Page 3 / 6

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.

rectangular field solutions
What is this?
Donna
the proudact of 3x^3-5×^2+3 and 2x^2+5x-4 in z7[x]/ is
?
Choli
a rock is thrown directly upward with an initial velocity of 96feet per second from a cliff 190 feet above a beach. The hight of tha rock above the beach after t second is given by the equation h=_16t^2+96t+190
Usman
Stella bought a dinette set on sale for $725. The original price was$1,299. To the nearest tenth of a percent, what was the rate of discount?
44.19%
Scott
40.22%
Terence
44.2%
Orlando
I don't know
Donna
if you want the discounted price subtract $725 from$1299. then divide the answer by $1299. you get 0.4419... but as percent you get 44.19... but to the nearest tenth... round .19 to .2 and you get 44.2% Orlando you could also just divide$725/$1299 and then subtract it from 1. then you get the same answer. Orlando p mulripied-5 and add 30 to it Tausif Reply p mulripied-5 and add30 Tausif p mulripied-5 and addto30 Tausif Can you explain further Monica Reply p mulripied-5 and add to 30 Tausif How do you find divisible numbers without a calculator? Jacob Reply TAKE OFF THE LAST DIGIT AND MULTIPLY IT 9. SUBTRACT IT THE DIGITS YOU HAVE LEFT. IF THE ANSWER DIVIDES BY 13(OR IS ZERO), THEN YOUR ORIGINAL NUMBER WILL ALSO DIVIDE BY 13!IS DIVISIBLE BY 13 BAINAMA When she graduates college, Linda will owe$43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was $1,585. What is the amount of each loan? Ariana Reply Sean took the bus from Seattle to Boise, a distance of 506 miles. If the trip took 7 2/3 hours, what was the speed of the bus? Kirisma Reply 66miles/hour snigdha How did you work it out? Esther s=mi/hr 2/3~0.67 s=506mi/7.67hr = ~66 mi/hr Orlando hello, I have algebra phobia. Subtracting negative numbers always seem to get me confused. Alicia Reply what do you need help in? Felix subtracting a negative....is adding!! Heather look at the numbers if they have different signs, it's like subtracting....but you keep the sign of the largest number... Felix for example.... -19 + 7.... different signs...subtract.... 12 keep the sign of the "largest" number 19 is bigger than 7.... 19 has the negative sign... Therefore, -12 is your answer... Felix —12 Niazmohammad Thanks Felix.l also get confused with signs. Esther Thank you for this Shatey ty Graham think about it like you lost$19 (-19), then found $7(+7). Totally you lost just$12 (-12)
Annushka
I used to struggle a lot with negative numbers and math in general what I typically do is look at it in terms of money I have -$5 in my account I then take out 5 more dollars how much do I have in my account well-$10 ... I also for a long time would draw it out on a number line to visualize it
Meg
practicing with smaller numbers to understand then working with larger numbers helps too and the song/rhyme same sign add and keep opposite signs subtract keep the sign of the bigger # then you'll be exact
Meg
Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles
168.50=R
Heather
john is 5years older than wanjiru.the sum of their years is27years.what is the age of each
46
mustee
j 17 w 11
Joseph
john is 16. wanjiru is 11.
Felix
27-5=22 22÷2=11 11+5=16
Joyce
I don't see where the answers are.
Ed
Cindy and Richard leave their dorm in Charleston at the same time. Cindy rides her bicycle north at a speed of 18 miles per hour. Richard rides his bicycle south at a speed of 14 miles per hour. How long will it take them to be 96 miles apart?
3
Christopher
18t+14t=96 32t=96 32/96 3
Christopher
show that a^n-b^2n is divisible by a-b
What does 3 times your weight right now
Use algebra to combine 39×5 and the half sum of travel of 59+30
Cherokee
What is the segment of 13? Explain
Cherokee
my weight is 49. So 3 times is 147
Cherokee
kg to lbs you goin to convert 2.2 or one if the same unit your going to time your body weight by 3. example if my body weight is 210lb. what would be my weight if I was 3 times as much in kg. that's you do 210 x3 = 630lb. then 630 x 2.2= .... hope this helps
tyler
How to convert grams to pounds?
paul