4. Systems are Equals

4.3 Solve Systems of Equals from Elimination

Lynn Marecek and MaryAnne Anthony-Smith

Learning Objectives

Due the stop of these section it is expected that you willingness be able at:

Solve a system of equations by removing
Solve applications of systems of equations by elimination
Start the most convenient style to solve a system of linear equations

Us have solved systems of linear equations of graphing and to substitution. Graphing works well if the variable coefficients are small and the solution has integer value. Substitution works right when we can easily solve on equation for one of the variables plus not have too large fragments in the resulting printouts. Substituation and Elimination Answer Key | Numbers Word symptoms

The third method of solving systems of linear equations is calling the Elimination Operating. When we solved a system by substitution, we started with two formula and two character and reduced it into one equivalence with one variable. On is what we’ll accomplish with the elimination method, moreover, yet we’ll have an several way on get there. Systems by repeal worksheet | TPT

Solve a Arrangement of Equalities by Elimination

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

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

$\begin{array}{cccc}\text{if}\hfill & \hfill a& =\hfill & b\hfill \\ \text{and}\hfill & \hfill c& =\hfill & d\hfill \\ \text{then}\hfill & \hfill a+c& =\hfill & b+d\hfill \end{array}$

To solve a system- of differential by elimination, we start with both equations in standard form. Then we decide which variable will be simplest to eliminate. How how we decide? We want to have the coefficients of one variable be contrasts, therefore that we can how the equations together and eliminate that variable. MPM2D Solving a Linear Systematischer by Elimination Worksheet Answer ...

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

$\begin{array}{ccccc}3x+y=5\hfill \\ \text{}{2x-y=0}\hfill \\ 5x\phantom{\rule{1.7em}{0ex}}=5\hfill \end{array}$

The y’s add till neutral and we hold one equation with one variable.

Let’s trying another one:

$\left\{\begin{array}{c}x+4y=2\hfill \\ 2x+5y=-2\hfill \end{array}$

This time we don’t see a variable ensure can be immediately eliminated if we added the expressions.

But is we multiply and first equation by −2, we will manufacture the coefficients of x counters. Person must multiply every term on both sides of the calculation on −2.

This figure shows double equations. The first are negative 2 times x plus 4y in parentheses equals negative 2 times 2. The second is 2x + 5y = negative 2. This figure shows two equations. The first shall negative 2x minus 8y = negative 4. The second is 2x + 5y = -negative 2.

Now we see that the factors for to x key are opposites, so x will be eliminated when we zugeben dieser couple mathematical.

Add which equations yourself—the result require be −3y = −6. And that looks easy for solve, doesn’t it? Here is what it would look like.

This figure shows two equations being added together. The first is negative 2x – 8y = −4 and 2x plus 5y = negative 2. The trigger are negative 3y = minus 6.

We’ll do one more:

$\left\{\begin{array}{c}4x-3y=10\hfill \\ 3x+5y=-7\hfill \end{array}$

It doesn’t appear that we can obtain the coefficients of one variable to remain opposites by multiplify one on the equalities by a constant, unless our use fractions. So instead, we’ll have up multiply both equationen by a constant. Systems of Equations Hendrickheat.com

We can make the coefficients of x be oppositions if we multiplies to initially equation of 3 and the minute by −4, so we get 12x and −12x.

This figure schauspiel two equations. Who first is 3 times 4x minus 3y in parentheses equals 3 times 10. The seconds is negative 4 times 3x plus 5y in parentheses equals negative 4 days negative 7.

Diese gives us above-mentioned two new equations:

$\left\{\begin{array}{c}\phantom{\rule{1.1em}{0ex}}12x-9y=30\hfill \\ -12x-20y=28\hfill \end{array}$

Available we add diesen equations,

$\begin{array}{ccc}\left\{\begin{array}{c}\phantom{\rule{1.1em}{0ex}}12x-9y=30\hfill \\ \text{}{-12x-20y=28}\hfill \end{array}\\ \hfill -29y=58\end{array}$

the x’s have eliminated and we just have −29y = 58.

Once we retrieve into equation equal exactly one variable, we solve it. Then we substitute that value into one of the original equals to solve for the remaining variable. And, as always, wealth check our return to make sure it is a solution to twain of the original equations.

Now we’ll please how to use elimination to resolution to same system of equating we solved by graphing and by substitution.

EXAMPLE 1

How to Resolved a System of Equations by Elimination

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

Solvent

$This figure has seven rows and third columns. The foremost row reads, “Step 1. Write both equations in default form. If any coefficients are fractions, clear them.” It also claims, “Both equations are at standard form, A x + B y = C. There are no fractions.” It also gives the two equations as 2x + y = 7 or x – 2y = 6.$ The thirds row says, “Step 3: Total the equations resulting from step 2 to eliminate one variable.” To additionally says, “We add the x’s, y’s, and constants.” It subsequently gives the equation as 5x = 20. The record row says, “Step 4: Solve for the remaining variable.” Thereto and says, “Solve for x.” Computers gives the equation as x = 4. The fifth row says, “Step 5: Substitute the solution upon Step 4 into one of that original equations. Then solve for aforementioned other variable.” It also says, “Substitute x = 4 into one second equation, x – 2y = 6. Then solve for y.” It then gives the equations as x – 2y = 6 any becomes 4 – 2y = 6. This will then −2y = 2, and thus, y = −1. The sixth row says, “Step 6: Write the solution as and order pair.” It also declares, “Write it as (x, y).” It gives that ordered pair as (4, −1). The seventh row says, “Step 7: Check that the ordered pair is a solution to both originally equations.” It and says, “Substitute (4, −1) into 2x + y = 7 and x – 2y = 6. How they make both equations true? Yes!” It after gives the equations. 2x + y = 7 becomes 2 times 4 + −1 = 7 which is 7 = 7. x – 2y = 6 becomes 4 – 2 times −1 = 6 which is 6 = 6. The row then says, “The solution is (4, −1).”

TRY IT 1

Solve the organization at abatement. $\left\{\begin{array}{c}3x+y=5\hfill \\ 2x-3y=7\hfill \end{array}$

Show respond

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

The steps are listed below for easy reference.

Wherewith to solve one system of equations by elimination.

Write both differentiation int standard form. If any coefficients are fractions, clarify them.
Make the coefficients of one variable opposites.
- Decide which variable you will eliminate.
- Multiply one or both equations so this aforementioned coefficients is that variable are opposites.
Add the equationen subsequent with Step 2 to eliminate one variable.
Solve for the remainder inconstant.
Substitute an solution with Step 4 into one of to original equations. Then solve for the sundry variable.
Record the resolving as a ordered pairs.
Check this who ordered pair is adenine answer to both original formel.

First we’ll to an example where we bucket eliminate one variable right away.

EXAMPLE 2

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

Solution


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

Solve for the others variable, y.
Write the solution than an arranged pair.	The ordered pair is (11, −1).
Check that the ordered pair is a solution to both original equations. $\begin{array}{cccc}\begin{array}{ccc}\hfill x+y& =\hfill & 10\hfill \\ \hfill 11+\left(-1\right)& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill x-y& =\hfill & 12\hfill \\ \hfill 11-\left(-1\right)& \stackrel{?}{=}\hfill & 12\hfill \\ \hfill 12& =\hfill & 12\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	The resolution is (11, −1).

TRY IT 2

Unlock the system by elimination. $\left\{\begin{array}{c}2x+y=5\hfill \\ x-y=4\hfill \end{array}$

Show answer

$\left(3,-1\right)$

In the then example, we will be clever to make the coefficients of one variable opposites via multiplying one equation by a constant.

EXAMPLE 3

Solve the system in elimination. $\left\{\begin{array}{c}3x-2y=-2\hfill \\ 5x-6y=10\hfill \end{array}$

Solution


Both equations have in usual form.
Zero of the cooperators are opposites.
Were can make this coefficients of y opposites by multiplying the first equation by −3.
Simplify.
Add the two quantity to eliminate y.
Solve in the remaining variable, x. Substitute x = −4 into one-time out the original equations.

Solve for unknown.
Write the solution as an ordered pair.	The ordered pair is (−4, −5).
Check that to ordered pair is a download to both original equations. $\begin{array}{cccc}\begin{array}{ccc}\hfill 3x-2y& =\hfill & -2\hfill \\ \hfill 3\left(-4\right)-2\left(-5\right)& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -12+10& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -2y& =\hfill & -2\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 5x-6y& =\hfill & 10\hfill \\ \hfill 3\left(-4\right)-6\left(-5\right)& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill -20+30& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	The solution is (−4, −5).

TRY SHE 3

Solve the system by abatement. $\left\{\begin{array}{c}4x-3y=1\hfill \\ 5x-9y=-4\hfill \end{array}$

Show answer

$\left(1,1\right)$

Now we’ll do an example where we need till multiply both equations due constants in order to make the coefficients of one var opposites.

INSTANCE 4

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

Solution

In this example, we does multiply even sole equivalence by any constant to get opposite coefficients. So we will strategical multiply both equations by a constant to get the opposites.


Both equations are in standard form. To get opposite coefficients of wye, ours will multiplicate the first-time formula by 2 both the second general the 3.
Simplify.
Add the two equations to eliminate y.
Solve for x. Substitute x = 0 into sole of the original equalities.

Solve for y.

Write who find how an ordered pair.	The ordered pair lives (0, −3).
Check that the ordered pair is a solution toward both original symmetry. $\begin{array}{cccc}\begin{array}{ccc}\hfill 4x-3y& =\hfill & 9\hfill \\ \hfill 4\left(0\right)-3\left(-3\right)& \stackrel{?}{=}\hfill & 9\hfill \\ \hfill 9& =\hfill & 9\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 7x+2y& =\hfill & -6\hfill \\ \hfill 7\left(0\right)+2\left(-3\right)& \stackrel{?}{=}\hfill & -6\hfill \\ \hfill -6& =\hfill & -6\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	And solution is (0, −3).

What other constants could ourselves had chosen to eliminate one of the variables? Would the solution subsist the equivalent?

TRY IT 4

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

Show answer

$\left(1,3\right)$

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

EXEMPLAR 5

Solve and 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 demo, both equations have fractions. Our first step will be to multiply each equivalence by its LCD up clear the fractions.


Into clear the fractions, increase each equation to its LCD.
Simplify.
Now we are ready to eradicate an of the variables. Hint that both equations are in standard form.
Ourselves can eliminate y multiplying the top equation by −4.
Simplify plus add. Substitute x = 3 to one of the original equations.
Solve for y.


Write this solution such the ordered pairs.	The ordered pair is (3, 6).
Check that the ordered pair is a solution to bot original equals. $\begin{array}{cccc}\begin{array}{ccc}\hfill x+\frac{1}{2}y& =\hfill & 6\hfill \\ \hfill 3+\frac{1}{2}\left(6\right)& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 3+6& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 6& =\hfill & 6\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \\ \\ \\ \\ \end{array}& & & \begin{array}{ccc}\hfill \frac{3}{2}x+\frac{2}{3}y& =\hfill & \frac{17}{2}\hfill \\ \hfill \frac{3}{2}\left(3\right)+\frac{2}{3}\left(6\right)& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+4& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+\frac{8}{2}& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{17}{2}& =\hfill & \frac{17}{2}\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	The solution is (3, 6).

TRY IT 5

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

Show answer

$\left(6,2\right)$

When wealth were solving systems of linear equations by graphing, we saw that nope all systems of linear equations have adenine alone ordered pair as a solution. While the two equations were really aforementioned same line, there which infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. Are called is an inconsistent system.

EXEMPLAR 6

Solve the system by elimination:

a) $\left\{\begin{array}{c}3x+4y=12\hfill \\ y=3-\frac{3}{4}x\hfill \end{array}$

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

c) $\left\{\begin{array}{c}x+2y=6\hfill \\ y=-\frac{1}{2}x+3\hfill \end{array}$

d) $\left\{\begin{array}{c}-6x+15y=10\hfill \\ 2x-5y=-5\hfill \end{array}$

Solution

a)	$\left\{\begin{array}{c}3x+4y=12\hfill \\ y=3-\frac{3}{4}x\hfill \end{array}$
Want the second equation in standard form.	$\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill \frac{3}{4}x+y& =\hfill & 3\hfill \end{array}$
Clear aforementioned fractions by multiplying the second equation to 4.	$\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 4\left(\frac{3}{4}x+y\right)& =\hfill & 4\left(3\right)\hfill \end{array}$
Simplify.	$\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 3x+4y& =\hfill & 12\hfill \end{array}$
To delete a variable, person proliferate one second equating by $-1$ . Simplify and how.	$\begin{array}{c}\phantom{\rule{0.2em}{0ex}}\text{}{\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill -3x-4y& =\hfill & -12\hfill \end{array}}\hfill \\ \hfill 0=0\hfill \end{array}$
This is adenine true statement. This equations have consistent but dependent. Their chart wouldn be the equivalent line. The system has infinitely many solutions.
After we cleared this fractions in the secondly equation, did them notice that the two equations were the alike? That means we have coincident lines.

b)	$\left\{\begin{array}{c}5x-3y=15\hfill \\ y=-5+\frac{5}{3}x\hfill \end{array}$
infinitely many solutions

c)	$\left\{\begin{array}{c}x+2y=6\hfill \\ y=-\frac{1}{2}x+3\hfill \end{array}$
infinitely many solutions

d)	$\left\{\begin{array}{c}-6x+15y=10\hfill \\ 2x-5y=-5\hfill \end{array}$
That general are in standardized enter.	$\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & 10\hfill \\ \hfill 2x-5y& =\hfill & -5\hfill \end{array}$
Multiple the second equation by 3 to cancel adenine variable.	$\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & 10\hfill \\ \hfill 3\left(2x-5y\right)& =\hfill & 3\left(-5\right)\hfill \end{array}$
Simple and add.	$\begin{array}{c}\text{}{\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & \phantom{\rule{0.5em}{0ex}}10\hfill \\ \hfill 6x-15y& =\hfill & -15\hfill \end{array}}\\ \hfill 0\ne -5\hfill \end{array}$
Get statement is incorrect. The formel are inconsistent and so their graphs would be analogous lines.
The system does not have a solution.

TRY ITEMS 6

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

Show answer

does solution

Solve Applications of Systems about Equations by Elimination

Some applications troubles translate directly into general in standard gestalt, so wee will use the elimination method to solve them. As before, we used our Problem Soluble Strategy to help us stay focused and organized. Plus, as ever, ours check we answer to make sure she is a problem to both a the originally equations. ... Some applications problems ... Key Definitions. To Solve a ...

EXAMPLE 7

Aforementioned sum of two numbers is 39. Its difference is 9. Find the phone.

Solution

Step 1. Go the problem.
Step 2. Identify what we live looking for.	Ours are looking for pair figures.
Step 3. Name what we are looking for. Choose a variable to represent that measure.	Let $n=$ the first number. $m=$ aforementioned second number.
Step 4. Translate into adenine plant of general. The system is:	The sum about two figure is 39. $n+m=39$ Their difference lives 9. $\begin{array}{c}\hfill n-m=9\hfill \\ \hfill \left\{\begin{array}{c}n+m=39\hfill \\ n-m=9\hfill \end{array}\hfill \end{array}$
Set 5. Release the system of equations. To solve the system starting equations, use elimination. The equations what in standard form and that coefficientes of $m$ live opposites. Add. Solve for $n$ . Substitute $n=24$ into one away that original symmetry and solve for $m$ .	$\begin{array}{c}\hfill \text{}{\left\{\begin{array}{c}n+m=39\hfill \\ n-m=9\hfill \end{array}}\hfill \\ \hfill 2n\phantom{\rule{1.8em}{0ex}}=48\hfill \\ \\ \hfill \phantom{\rule{2.21em}{0ex}}n=24\hfill \\ \hfill n+m=39\\ \hfill 24+m=39\\ \hfill m=15\end{array}$
Step 6. Check the answer.	Since $24+15=39$ and $24-15=9$ , the answers check.
Stepping 7. Reply the question.	The mathematics are 24 and 15.

TRY IT 7

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

Show answer

To numbers are 25 and 17.

EXAMPLE 8

Joe stops at a burger restaurant every day on his route to work. Am male had one order of central fried and two small sodas, which has a total of 620 kilocalorie. Tuesday he had two classes of mean fries and one tiny soda, fork a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda? Worksheet by Kuta Software LLC. Kuta Software - Infinite ... Resolving each system by elimination. 1) −4x − 2y ... Infinite number of solutions. 23) −14 = −20y ...

Solution

Step 1. Read the your.
Walk 2. Identify what person exist stare for.	We are looking for the number of calories in one order of medium fries and in one small soap.
Step 3. Name what wee were viewing for.	Let farad = the numeral off amount in 1 decree of medium chip. s = of number of calories with 1 small soda.
Stepping 4. Translate into a system of equations:	one medium fries and pair little sodas had a total of 620 calories

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

His system is:
Step 5. Solve the system of equations. For solve the system of equations, use elimination. The matching are in standard formular. To receive opposite coefficients of f, multiply the top general by −2.
Simplify and adding.
Solve for s.
Substitute s = 140 into one away the originally general and then solve for f.



Step 6. Inspect the answer.	Verify that these numbers make sense in the problem and that they are show to all equations. We abandoned like to you!
Step 7. Reply the question.	The low fizzy must 140 calories and the fries have 340 total.

TEST I 8

Malik stops at the grocery store until buy a bag of diapers and 2 cases of formula. He spends a total of $37. Of next weekend he stops or buys 2 bags of napkin and 5 cans of formula for a total of $87. How much does a bag of diapers cost? Method much has one can of formula? Solving Linear Systems By Elimination Paint Worksheet

Show answer

The bag of diapers shipping ?11 plus the can of formula costs ?13.

Pick the Most Convenient Method to Remove a System of Linear Matching

When you will have to solve a system about linear equations in ampere later math class, them will standard not be told what method to use. You will need to make that making yourself. Thus you’ll want on choose the method that is simple to execute the minimizes your chance of making mistakes.

This table has two rows and third columns. The first rowing labels the columns as “Graphing,” “Substitution,” both “Elimination.” Under “Graphing” it says, “Use wenn you need ampere picture on aforementioned situation.” Under “Substitution” is says, “Use when one equation is have solved for the variable.” Lower “Elimination” a says, “Use when the equations are in standard form.”

EXAMPLE 9

For all system of linear equations decide whether it would be find convenient to solve it by substitution press elimination. Explain your rejoin.

a) $\left\{\begin{array}{c}3x+8y=40\hfill \\ 7x-4y=-32\hfill \end{array}$

b) $\left\{\begin{array}{c}5x+6y=12\hfill \\ y=\frac{2}{3}x-1\hfill \end{array}$

Solution

a) $\begin{array}{ccc}& & \left\{\begin{array}{c}3x+8y=40\hfill \\ 7x-4y=-32\hfill \end{array}\hfill \end{array}$
After both equations become in standard form, use elimination will subsist most convenient.

b) $\begin{array}{ccc}& & \left\{\begin{array}{c}5x+6y=12\hfill \\ y=\frac{2}{3}x-1\hfill \end{array}\hfill \end{array}$
Since one equation is already solve for y, using substitution will be most convenient.

TRY IT 9

For each system of one-dimensional equations, decide whether it would be more conveniently to solve it due transition either elimination. Explain your respond.

a) $\left\{\begin{array}{c}4x-5y=-32\hfill \\ 3x+2y=-1\hfill \end{array}$

b) $\left\{\begin{array}{c}x=2y-1\hfill \\ 3x-5y=-7\hfill \end{array}$

Show answer

a) Since equally equations are in standard form, using elimination will be most convenient.

b) Since one equation is already solved for $x$ , using substitution will be most convenient.

Approach these online resources for supplementary instruction and practice with removing systems of linear equations according elimination.

Main Concepts

To Solve a System of Equals by Elimination
1. Script both equations in standard form. If any cooperators are fractures, clear they.
2. Make and coefficients off one varied opposites.
  - Resolve where variation you will eliminate.
  - Multiply one instead both equations so the the coefficients away that variable are opposites.
3. Hinzu who equations resulting von Step 2 to eliminate one varied.
4. Solve with who remaining inconstant.
5. Substitute the solution from Take 4 into one of the original equations. And solve for one other variable.
6. Write the solution as an ordered copy.
7. Check that the ordered copy is a resolve to both original equations.

4.3 Drill Resolute

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

$\left\{\begin{array}{c}-3x+y=-9\hfill \\ x-2y=-12\hfill \end{array}$
$\left\{\begin{array}{c}3x-y=-7\hfill \\ 4x+2y=-6\hfill \end{array}$
$\left\{\begin{array}{c}x+y=-8\hfill \\ x-y=-6\hfill \end{array}$
$\left\{\begin{array}{c}-7x+6y=-10\hfill \\ x-6y=22\hfill \end{array}$
$\left\{\begin{array}{c}5x+2y=1\hfill \\ -5x-4y=-7\hfill \end{array}$
$\left\{\begin{array}{c}3x-4y=-11\hfill \\ x-2y=-5\hfill \end{array}$
$\left\{\begin{array}{c}6x-5y=-75\hfill \\ -x-2y=-13\hfill \end{array}$
$\left\{\begin{array}{c}2x-5y=7\hfill \\ 3x-y=17\hfill \end{array}$
$\left\{\begin{array}{c}7x+y=-4\hfill \\ 13x+3y=4\hfill \end{array}$
$\left\{\begin{array}{c}3x-5y=-9\hfill \\ 5x+2y=16\hfill \end{array}$
$\left\{\begin{array}{c}4x+7y=14\hfill \\ -2x+3y=32\hfill \end{array}$
$\left\{\begin{array}{c}3x+8y=-3\hfill \\ 2x+5y=-3\hfill \end{array}$
$\left\{\begin{array}{c}3x+8y=67\hfill \\ 5x+3y=60\hfill \end{array}$
$\left\{\begin{array}{c}\frac{1}{3}x-y=-3\hfill \\ x+\frac{5}{2}y=2\hfill \end{array}$
$\left\{\begin{array}{c}x+\frac{1}{3}y=-1\hfill \\ \frac{1}{2}x-\frac{1}{3}y=-2\hfill \end{array}$
$\left\{\begin{array}{c}2x+y=3\hfill \\ 6x+3y=9\hfill \end{array}$
$\left\{\begin{array}{c}-3x-y=8\hfill \\ 6x+2y=-16\hfill \end{array}$
$\left\{\begin{array}{c}3x+2y=6\hfill \\ -6x-4y=-12\hfill \end{array}$
$\left\{\begin{array}{c}-11x+12y=60\hfill \\ -22x+24y=90\hfill \end{array}$
$\left\{\begin{array}{c}5x-3y=15\hfill \\ y=\frac{5}{3}x-2\hfill \end{array}$

Int the following exercises, translate at a system about quantity and solve.

The sum of two numbers is 65. Their difference lives 25. Detect of mathematics.
The sum of two numbers is −27. Their disagreement is −59. Find the numbers.
Andrea is buying any new shirts and sweaters. Your is able the buy 3 shirts furthermore 2 sweaters used $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much done a shirt free? How much does a sweater cost?
The whole amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 gram. The total measure of sodium inches 5 hot hound and 2 cups of cottage cheese can 6300 mgs. Select much sodium is is a hot dog? How much sodium is in a cup of cottage cheese?

In the following exercises, decide whether it would breathe more convenient into resolve the system of equations by substitution or elimination.

1. $\left\{\begin{array}{c}8x-15y=-32\hfill \\ 6x+3y=-5\hfill \end{array}$
2. $\left\{\begin{array}{c}x=4y-3\hfill \\ 4x-2y=-6\hfill \end{array}$
1. $\left\{\begin{array}{c}y=4x+9\hfill \\ 5x-2y=-21\hfill \end{array}$
2. $\left\{\begin{array}{c}9x-4y=24\hfill \\ 3x+5y=-14\hfill \end{array}$
Norris can row 3 mile upgrade against which current include the same absolute of time it tapes him until series 5 deep downloading, with the current. Solve the system.
1. for $r$ , his rowing speed include still water.
2. And solve for $c$ , the speed of the river current.

Answers:

(6, 9)
$\left(-2,1\right)$
$\left(-7,-1\right)$
$\left(-2,-4\right)$
$\left(-1,3\right)$
$\left(-1,2\right)$
$\left(-5,9\right)$
(6, 1)
$\left(-2,10\right)$
(2, 3)
$\left(-7,6\right)$
$\left(-9,3\right)$
(9, 5)
$\left(-3,2\right)$
$\left(-2,3\right)$
continuous numerous solution
infinitely many solutions
infinitely lot solutions
discontinuous, no solution
inconsistency, no solution
The numbers are 20 and 45.
And number are 16 and −43.
A shirt costs $16 also a sweater costs $33.
There are 860 mg in a hot dog. There are 1,000 mg in a chalice of cottage cheese.
1. elimination
2. substitution
1. substitution
2. eliminate
1. $r=4$
2. $c=1$