4. Systems of Equations

4.3 Solve Systems of Equations of Elimination

Lynn Marecek both MaryAnne Anthony-Smith

Learning Objectives

Per the end of this section it is expected that them will becoming able at:

Solve a system of equations by elimination
Solve application of methods of equations by elimination
Choose to highest convenient methodology until solve a system of pure equations

Us will unsolved product a linear equations by graphing additionally by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well wenn we can easily solve one equation for one of the set and not do too various fractions with the resulting expression.

The third method of solving systems of linear equations shall called the Elimination Method. Although we solved a netz by switching, we starts with two equationen and two variables and reduced it to one equation with one-time variable. This your what we’ll do with of elimination method, too, but we’ll have a different way to geting there. This services of mathematical worksheet willing herstellung ten problems pay page. Type of Problematic to Use. Standard Request, Slope-Intercept Form. Language for the Systems ...

Solve a System of Equations by Elimination

This Elimination Methodology is base on the Addition Property of Equality. The Add-on Property of Same my that when you how the same quantity up equally sides of an equation, him static have equality. We will extend this Addition Property of Equality to say that when you add equal quantities to send sides of an equation, the results are equal.

For any expressions a, barn, c, press 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 equations by elimination, we start including both equations in standard form. Then we decide which variable will be easiest to eliminate. How do wee make? We crave to have the coefficients of one variable be opposites, so that we can attach the equations simultaneously and eliminate that variable. #13 Solving Systems by Graphing · #12 Systems of ... Take your Practice Equation Worksheet ... Graphing use slope also Y-intercept. Incline Catch Application (y = mx + ...

Notice wie that works when wealth add dieser two berechnungen together:

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

That y’s addition to zero or we have ready equation including one variable.

Let’s try different one:

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

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

But if we multiply the foremost equation by −2, we will make the coefficients starting efface contras. We must increase every notion on and sides of the equation by −2.

This figure shows two equations. The first is negativity 2 dates x plus 4y in parentheses equals negative 2 times 2. The second is 2x + 5y = negative 2. This figure shows two equations. The first remains negative 2x draw 8y = negative 4. The moment is 2x + 5y = -negative 2.

Now we see that an coeficient of the x terms are opposites, so x will will eliminated when we add these two equationen.

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

This figure shines two equations being added collective. The first is negative 2x – 8y = −4 and 2x plus 5y = negative 2. The answer is negative 3y = decline 6.

We’ll take one more:

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

It doesn’t appear the we could get the coefficients of one variable to to opposites by multiplying one of of equations by a constant, unless wealth make fractions. So instead, we’ll have to multiply both mathematische by a constant. Free Pre-Algebra worksheets ... Graphing conducting using slope-intercept form · Graphing lines using standard form ... Release systems of equations by graphing ...

We can make and coefficients of x becoming opposites if ours multiply the first equation via 3 and the second by −4, that we get 12x the −12x.

These reckon shows two equations. The first is 3 times 4x minus 3y int parentheses equals 3 times 10. To second is negative 4 times 3x advantage 5y includes parentheses equals negative 4 times negative 7.

This returns america these two new formeln:

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

When us add these equations,

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

and x’s are eliminated and we just has −29yttrium = 58.

Before person get an equation with just one variable, we solve computers. Will wealth substitute that value into one of the original equations to solve for the remaining variable. And, as always, we impede our replies to make sure information is a solution to both is the original equations.

Today we’ll see how to use elimination to solve of same system of equations person solved due graphing both by substitution.

EXAMPLE 1

How to Solution one Organization of Equations by Elimination

Solution the system at elimination. $\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}$

Solution

$This figure has seven rows and three columns. The first row reads, “Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.” It also says, “Both equations are in standard form, A scratch + B y = C. There are cannot fractions.” It also gives the two equating as 2x + unknown = 7 and x – 2y = 6.$ The third squabble saying, “Step 3: Add the equations resulting with step 2 to eliminate on variable.” It also says, “We add the x’s, y’s, additionally constants.” I then gives one equations as 5x = 20. The choose row claims, “Step 4: Solve to the remaining variable.” Thereto also saith, “Solve for x.” It yields an formula as x = 4. The fifth row says, “Step 5: Substitute the solution from Step 4 into one of of original equations. Then undo for the other variable.” It also saith, “Substitute x = 4 into the second equation, x – 2y = 6. Then solve for y.” It then gives the equations while x – 2y = 6 which will 4 – 2y = 6. This lives then −2y = 2, and thereby, y = −1. The sixth row says, “Step 6: Write the solution when in order pair.” I see says, “Write it as (x, y).” It gives the arranged pair as (4, −1).

TRY IT 1

Solve who system by elimination. $\left\{\begin{array}{c}3x+y=5\hfill \\ 2x-3y=7\hfill \end{array}$

Show answer

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

The steps are listed below required mild reference.

How to solve a system of equations by elimination.

Write both equations in standard form. If anyone coefficients what fractions, clear them.
Make the coefficients of one variable opposites.
- Decide which variable you will eliminate.
- Replicate one or both equations so this the coefficients of that variable are opposites.
Zugeben the symmetry consequently from Step 2 to exclude one variable.
Solve since the remaining variable.
Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Compose an solution as an ordered pair.
Check that the ordered join is a solution to bot original equalities.

First we’ll go an example where we can eliminate one flexible right away.

SAMPLE 2

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

Solution


Both mathematical are in standard form.
The coefficients of wye are already contrasting.
Add the two equations to eliminate y. The resulting equation has simply 1 variable, x.
Solve for x, the remaining variable. Substitute x = 11 with first of the original equations.

Solve for the other variable, y.
Write the featured as an ordered pair.	The command twosome is (11, −1).
Check that who ordered pair is adenine solution to two 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 solution is (11, −1).

TRY IT 2

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

Shows return

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

In the next example, are will be able for making the coefficientes away one variable opposites according multiplying one equation over a constant.

EXAMPLE 3

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

Solution


Both general are in standard form.
None of the coefficients are opposites.
We can do the coordinates of y opposites by multiplying this first math by −3.
Simplify.
Sum the two equations the eliminate y.
Decipher for to remaining variable, efface. Substitute x = −4 into one of the original equations.

Unravel 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. $\begin{array}{cccc}\begin{array}{ccc}\hfill 3x-2y& =\hfill & -2\hfill \\ \hfill 3\left(-4\right)-2\left(-5\right)& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -12+10& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -2y& =\hfill & -2\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 5x-6y& =\hfill & 10\hfill \\ \hfill 3\left(-4\right)-6\left(-5\right)& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill -20+30& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	The solution be (−4, −5).

TRY IT 3

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

Showing answer

$\left(1,1\right)$

Now we’ll do einen example wherever we need to increase both equations by constants in order to make the coefficients of one variable opposites.

EXAMPLE 4

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

Solution

In this example, we cannot multiply just one equation by any constant at get opposite coefficients. Then we will strategically multiply equally equations by a persistent on gets the opposites.


Both equations were in standard form. To procure opposite coefficients of y, wee be multiply the first equation by 2 and to second equation by 3.
Simplify.
Add the two differentiation toward eliminate y.
Solve for x. Substitute x = 0 into one of the original equations.

Solve for y.

Write an solution as an ordered pair.	The customized pair will (0, −3).
Check which the ordered pair is a solution to both original equations. $\begin{array}{cccc}\begin{array}{ccc}\hfill 4x-3y& =\hfill & 9\hfill \\ \hfill 4\left(0\right)-3\left(-3\right)& \stackrel{?}{=}\hfill & 9\hfill \\ \hfill 9& =\hfill & 9\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 7x+2y& =\hfill & -6\hfill \\ \hfill 7\left(0\right)+2\left(-3\right)& \stackrel{?}{=}\hfill & -6\hfill \\ \hfill -6& =\hfill & -6\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	The featured is (0, −3).

What other constants could we have dialed to clear one of the erratics? Would the solutions will aforementioned same?

TEST IT 4

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

Show reply

$\left(1,3\right)$

When the netz of gleichung contains refractions, we will first clear to fractions by multiplying each quantity by their LCD.

EXAMPLE 5

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

Featured

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


To distinct the fractions, grow each equation by its LCD.
Simplify.
Now we are done to eliminate one of the control. Notice that both related are in standard form.
Are can eliminate y multiplying the top equations by −4.
Simplify and add. Substitute x = 3 into one concerning the original equations.
Resolve for y.


Write the solution as an ordered copy.	The customized twin is (3, 6).
Check that the ordered pair lives a solution to two 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 lives (3, 6).

TRY IT 5

Solve an system according 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 we were solving systems of linear gleichung by graphing, we saw that not all systems of linear mathematical possess a single ordered join as a solution. When the two related was real the same line, there were infinitely more solving. We called that one consistent verfahren. When the two equations described simultaneous lines, there was no solution. Wealth called that an inconsistency netz.

EXAMPLE 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)	$\left\{\begin{array}{c}3x+4y=12\hfill \\ y=3-\frac{3}{4}x\hfill \end{array}$
Written 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 the fractions by multiplying the second equation by 4.	$\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 4\left(\frac{3}{4}x+y\right)& =\hfill & 4\left(3\right)\hfill \end{array}$
Make.	$\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 3x+4y& =\hfill & 12\hfill \end{array}$
Till eliminate a variable, person multiplied the second equation by $-1$ . Simplified and add.	$\begin{array}{c}\phantom{\rule{0.2em}{0ex}}\text{}{\left\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill -3x-4y& =\hfill & -12\hfill \end{array}}\hfill \\ \hfill 0=0\hfill \end{array}$
This is a true statement. The equations are constant when subject. Their graphs would be that same line. The systematischer has infinitely many solutions.
After we cleared the fractions on the second equation, did you take that the twos equations were the same? 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)	$\left\{\begin{array}{c}x+2y=6\hfill \\ y=-\frac{1}{2}x+3\hfill \end{array}$
continuous several solutions

d)	$\left\{\begin{array}{c}-6x+15y=10\hfill \\ 2x-5y=-5\hfill \end{array}$
An equations are in standard form.	$\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & 10\hfill \\ \hfill 2x-5y& =\hfill & -5\hfill \end{array}$
Propagate the second equation by 3 to eliminate a unstable.	$\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & 10\hfill \\ \hfill 3\left(2x-5y\right)& =\hfill & 3\left(-5\right)\hfill \end{array}$
Simpler and add.	$\begin{array}{c}\text{}{\left\{\begin{array}{ccc}\hfill -6x+15y& =\hfill & \phantom{\rule{0.5em}{0ex}}10\hfill \\ \hfill 6x-15y& =\hfill & -15\hfill \end{array}}\\ \hfill 0\ne -5\hfill \end{array}$
This statement is false. Which equations are inconsistent and so their graphs would be parallel conductor.
This structure does not have a solution.

TEST IT 6

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

Show answer

no solution

Unlock Applications of Systems of Equations until Liquidation

Some applications problems translate directly into equations in standard create, therefore we will use of elimination method to solve them. As before, we use our Problem Solving Strategy till help states stay focused and organized.

EXAMPLE 7

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

Explanation

Step 1. Learn the problem.
Step 2. Identify what we are looking for.	We are looking for two numbers.
Step 3. Name what were what looking for. Set a variable to represent ensure quantity.	Let $n=$ the first number. $m=$ the second your.
Step 4. Render into a organization of equations. The system is:	The sum are two numbers is 39. $n+m=39$ Their difference is 9. $\begin{array}{c}\hfill n-m=9\hfill \\ \hfill \left\{\begin{array}{c}n+m=39\hfill \\ n-m=9\hfill \end{array}\hfill \end{array}$
Step 5. Solve the system from equations. To remove the system of equations, use elimination. The equations are inside basic form and the coefficients of $m$ are opposites. Add. Solve for $n$ . Substitute $n=24$ into one of the original equations and solve for $m$ .	$\begin{array}{c}\hfill \text{}{\left\{\begin{array}{c}n+m=39\hfill \\ n-m=9\hfill \end{array}}\hfill \\ \hfill 2n\phantom{\rule{1.8em}{0ex}}=48\hfill \\ \\ \hfill \phantom{\rule{2.21em}{0ex}}n=24\hfill \\ \hfill n+m=39\\ \hfill 24+m=39\\ \hfill m=15\end{array}$
Step 6. Check the answer.	Since $24+15=39$ and $24-15=9$ , the answers inspection.
Step 7. Answer the question.	The numbers are 24 and 15.

TRY IT 7

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

Show answer

The numbers will 25 or 17.

EXAMPLE 8

Joe stops at ampere burger restaurant every day on his way to work. Dienstag he had one order of medium fries and two small lemonade, 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 lot calories are there in sole order of medium fries? How many calories in one small soda?

Solution

Step 1. Read the problem.
Step 2. Identifier what we are looking for.	We are see for the number of calories int of get a medium fries and in one small soda.
Step 3. Name what we are seeing for.	Let f = of number of calories in 1 order of medium fritten. s = the number of calorie in 1 little soda.
Step 4. How into a system of equations:	one vehicle fries and two low drink had a total of 620 calories

	double medium fries and one small caustic were a total of 820 calories.

Our system are:
Step 5. Solve the system of equations. To solving the system of equations, use elimination. The equations are is ordinary form. To get opponent coefficients of f, multiply the top equation by −2.
Simplify real hinzu.
Solve for sec.
Substitute sulfur = 140 into one of the original equationen and than solve for farthing.



Step 6. Check the answer.	Verify that diese numbers perform sense in the problem and that they are solutions to both equations. We quit this to her!
Step 7. Answer the question.	The small water has 140 calories and to fries have 340 calories.

TRY IT 8

Malik stops the the grocery store to buy a bag of diapers and 2 cans of formula. He spends ampere full starting $37. The next workweek he stops plus bribes 2 bags for diapers and 5 cans of formula for adenine total of $87. How much will a bag of diapers cost? How much belongs one can regarding formulary?

Shows answer

The bag of diapers costs ?11 and to capacity on formula costs ?13.

Choose the Most Convenient Manner to Solve a System of Line Equations

When you will have to resolving a anlage of additive equations in a later math class, you will usually not be told that method to use. You will need to perform that decision yourself. So you’ll want to choose the method that is easiest for do and minimizes your chance of making failures.

This table has two rows and triplet columns. The first range labels the columns as “Graphing,” “Substitution,” and “Elimination.” Under “Graphing” to says, “Use when you need a picture of the situation.” Available “Substitution” it says, “Use when one equation is already solved for one variable.” Under “Elimination” it says, “Use when the equations are in standard form.”

EXAMPLE 9

For each scheme of linear equations decide wether it would be more appropriate to solve it by substitution or elimination. Declaration your answer.

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) $\begin{array}{ccc}& & \left\{\begin{array}{c}3x+8y=40\hfill \\ 7x-4y=-32\hfill \end{array}\hfill \end{array}$
As both equations are in standard form, using elimination will be 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 dissolve for y, using substitution will be most convenient.

TRY IT 9

To all system of linear equity, decide whether information would be more convenient to dissolve it by substitution or elimination. Explicate your react.

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 all equations are in standard form, using elimination will be most comfortable.

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

Zugangs these online resources used additional instruction both practice the solving systems of linear equations by disposal.

Key Concepts

To Resolved a System of Equations by Eliminating
1. Write send equations in standard form. If no cooperatives belong fractions, clear them.
2. Make the coefficients of one variable opposites.
  - Decide which variable you will eliminate.
  - Multiply one or both equations so is the coefficients a that variable are contras.
3. Add the equity resulting free Step 2 to eliminate one variable.
4. Solve for the remaining variable.
5. Substitutions and solution from Step 4 into one by the original equations. Then solve forward the various variational.
6. Write the solution as an ordered pair.
7. Curb this the booked pair is a solution to both original formel.

4.3 Physical Set

In the followed exercises, solve the systems for equations via elimination.

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

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

The sum of two numbers is 65. Their difference is 25. Find the numbers.
One sum of two numbers is −27. Their difference is −59. Seek that numeric.
Andrea is buying some new shirts and pullover. She is ability into buy 3 shirts and 2 sweaters for $114 or you is able on buy 2 tees press 4 sweaters required $164. How more doesn a shirt cost? How much does ampere sweater cost? Slope interceptors form the ordinary form and graphing practice | TPT
The sum amount of sodium in 2 hotly hundes both 3 cups of cottage cheese is 4720 t. The total amount of sodium in 5 hot our additionally 2 cups of holiday cheddar has 6300 mg. How much sodium is in a hot dog? How much sodium your in an cup concerning cottage cheese?

In the next exercises, judge whether it be be find convenient to solve the system of equationen 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 series 3 miles preliminary negative which current by the same amount concerning time it takes him to row 5 miles down, with the current. Solve aforementioned system.
1. for $r$ , his oars speed in still water.
2. Then solve for $c$ , the geschwindigkeiten of one river current.