3. Solving Systems of Equations by Elimination
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Chapter 5
3.
Solving Systems of Equations by Elimination
| Student Learning Objectives: |
|---|
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Why
This lesson explores the elimination method as a technique for solving systems of equations. It emphasizes the importance of understanding the terms
consistencyand
independencein the context of these systems. These concepts help in determining whether a system has one solution, no solutions, or infinitely many solutions. Real-world applications include budgeting for school gym equipment, calculating the number of toys in a box, and determining the profit from selling tools at a garage sale. The elimination method is presented as a reliable and straightforward approach for solving these types of problems, making it a valuable skill for students and professionals alike.
Lesson Settings & Tools
| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Systems of Equations by Elimination
Slide of 11
The aim of this lesson is to show a method to solve a system of linear equations by adding or subtracting the equations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
How Many Toys Are There in Each Box?
Maya is researching a toy factory that currently produces robot action figures and car toys.
These toys are sold to stores in boxes. The total weight of the content of each box is of 20 kilograms. Each robot weighs 45 kilogram and each car weighs 110 kilogram. If x is the number of robots and y the number of cars in a box, the given information can be expressed by a linear equation. 4/5x + 1/10y = 20 Also, each robot and each car are sold to the stores for $15 and $5, respectively. The factory makes $500 for each box sold to a store. This information can also be written as an equation. 15x + 5y = 500 These equations can be combined to form a system of linear equations. 45x + 110y = 20 15x + 5y = 500 Solve this system of equations without graphing. How many robots and cars are there in each box?
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Discussion
Rewriting Equations
Sometimes the equations of a system can be rewritten in a way that is easier to work with. As an example, consider the following system of linear equations. x + 2y = 3 & (I) 12x - y = - 12 & (II) In Equation (II), the coefficient of the variable x is a fraction. It would be easier to solve the system if all the coefficients were integers. It is possible to rewrite the system using equivalent systems.
Rule
Equivalent Systems
In a system of equations, an equivalent system can be created by replacing one equation with the sum of two or more equations in the system or by replacing an equation with a multiple of itself. An equation can be also replaced by the sum of that equation and a multiple of another equation in the system.
Proof
The statement will be proved for linear equations. For non-linear equations, the proof is similar. Consider a system of linear equations.
ax+by=c & (I) dx+ey=f & (II)
Let ( x_1, y_1) be a solution to the system. Therefore, this ordered pair satisfies both equations simultaneously.
a x_1+b y_1=c ✓ d x_1+e y_1=f ✓
Consider now the system formed by the following two equations.
- Equation (I)
- Equation (II) plus a multiple of Equation (I)
The system described above is shown below. ax+by=c dx+ey+α (ax+by)=f+α c It is already known that ( x_1, y_1) is a solution to the first equation of this new system. It needs to be verified that the ordered pair is also a solution to the second equation. a x_1+b y_1=c ✓ d x_1+e y_1+α (a x_1+b y_1)? =f+α c Since ( x_1, y_1) is a solution to the original system, it is known that a x_1+b y_1=c and that d x_1+e y_1=f. These two expressions can be substituted into the second equation above.
Therefore, ( x_1, y_1) is also a solution to the second equation of the new system. This means that this ordered pair is a solution to the system formed by the first equation of the original system and the sum of the first equation and a multiple of the second equation. a x_1+b y_1=c ✓ d x_1+e y_1+α (a x_1+b y_1)=f+α c ✓
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Discussion
Elimination Method
Given a system of two equations in two variables, replacing one equation with the sum of that equation and a multiple of the other equation produces an equivalent system. This fact is used to solve systems of equations by the Elimination Method. Consider an example system of linear equations. 3x+2y=6 & (I) y=2x-11 & (II) To solve the system by using the Elimination Method, there are five steps to follow.
1
Write the Equations in the Same Form
expand_more First, all like terms must be gathered on the same sides of the equations. In Equation (I), the variable terms are on the same side of the equation. However, the variable terms are on both sides of the equations in Equation (II). Like terms can be gathered on the same sides of the equations by applying the Properties of Equality.
2
Multiply an Equation
expand_more Multiply one of the equations by a constant so that one of the variable terms of the resulting equation is equal to or is the opposite of the corresponding variable term in the other equation. In this case, multiplying Equation (II) by - 2 will produce opposite coefficients for the y-variable.
- 2 (- 2x+y)= - 2(- 11) ⇕ 4x-2y= 22
Both the original and the resulting equations have the same solutions because they are equivalent equations.
3
Add the New Equation and the Other Equation
expand_more After the rewrites, the system of equations looks the following way.
3x + 2y=6 4x - 2y=22
Add these two equations by adding the right-hand sides together and the left-hand sides together. This way one variable will be eliminated.
3x+2y+( 4x-2y)=6+ 22
▼
Simplify
RemovePar
Remove parentheses
3x+2y+4x-2y=6+22
CommutativePropAdd
Commutative Property of Addition
3x+4x+2y-2y=6+22
AddSubTerms
Add and subtract terms
7x= 28
Note that this step results in an equation in only one variable. This equation can be solved by dividing both sides by 7.
4
Write an Equivalent System
expand_more Substitute the value for the solved equation in one variable for any of the equations of the system. This produces an equivalent system of equations. In this case, Equation (I) will be replaced.
& 3x+2y=6 & (I) - 2x+y=- 11 & (II) & ⇕ & x= 4 - 2x+y=- 11
Note that the first equation is the solution value of x.
5
Solve the Equivalent System
expand_more To solve the new system, substitute the found value into the other equation. In this case, 4 will be substituted into Equation (II) for x to find the value of y.
In this system, the value of y is - 3. Therefore, the solution to the system of equations, which is the point of intersection of the lines, is (4,- 3), or x=4, y=- 3.
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Pop Quiz
Checking if the Coordinate Pair is a Solution
For each system of linear equations, verify whether the coordinate pair is a solution.
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Example
The Price of Gold
A vlogger that Diego likes to watch bought silver and gold to make Olympic-style medals. The vlogger will show the process of making the medals as a multi-video series.
The price of gold is about $2000 per ounce, while the price of of silver is about $25 per ounce. The vlogger spent a total of $7075 in both silver and gold. Let g and s be the ounces of gold and silver, respectively, that the vlogger bought. Then, the given information can be modeled by a linear equation. 25s + 2000g = 7075 Diego notes that if he subtracts 50 times the number of ounces of gold from 5 times the number of ounces of silver, there would still be 65 ounces left. This can also be written as an equation. 5s - 50g = 65 A system of equations is obtained by combining these two equations. 25s + 2000g = 7075 & (I) 5s - 50g = 65 & (II)
a Solve the system of equations graphically.
b Solve the system of equations using the Substitution Method.
c Solve the system of equations using the Elimination Method.
d Interpret the system in terms of consistency and independence.
Answer
a See solution.
b See solution.
c See solution.
d The system is consistent and independent.
Hint
a First write the equations in slope-intercept form.
b Isolate one variable in an equation.
c Recall that equivalent systems have the same solution.
d A system of equations can be inconsistent, consistent dependent, or consistent independent.
Solution
a To solve the system graphically, first each equation should be written in slope-intercept form. The variable s will arbitrarily be considered as the dependent variable, and g will be the independent variable.
25s + 2000g = 7075 & (I) 5s - 50g = 65 & (II)
▼
(I), (II): Write in slope-intercept form
DivEqn
(I): .LHS /25.=.RHS /25.
25s + 2000g25 = 283 5s - 50g = 65
WriteSumFrac
(I): Write as a sum of fractions
25s25 + 2000g25 = 283 5s - 50g = 65
DivEqn
(II): .LHS /5.=.RHS /5.
25s25 + 2000g25 = 283 5s - 50g5 = 13
WriteDiffFrac
(II): Write as a difference of fractions
25s25 + 2000g25 = 283 5s5 - 50g5 = 13
(I), (II): a* b/c=a/c* b
2525s + 200025g = 283 55s - 505g = 13
(I), (II): Calculate quotient
1s + 80g = 283 1s - 10g = 13
(I), (II): Identity Property of Multiplication
s + 80g = 283 s - 10g = 13
SubEqn
(I): LHS-80g=RHS-80g
s=- 80g+ 283 s - 10g = 13
AddEqn
(II): LHS+10g=RHS+10g
s=- 80g+ 283 s =10g+ 13
Now, both equations can be graphed on the same coordinate plane.
Looking at the graph, it can be seen that the solution is about 3 ounces of gold and a bit more than 40 ounces of silver. In this case, since the point of intersection is not a lattice point, finding the exact solution is not possible.
b To solve the system using the Substitution Method, first a variable should be isolated from one of the equations. It seems easier to isolate s on Equation (II).
25s + 2000g = 7075 & (I) 5s - 50g = 65 & (II)
▼
(II): Solve for s
DivEqn
(II): .LHS /5.=.RHS /5.
25s + 2000g = 7075 5s - 50g5 = 13
WriteDiffFrac
(II): Write as a difference of fractions
25s + 2000g = 7075 5s5 - 50g5 = 13
MovePartNumRight
(II): a* b/c=a/c* b
25s + 2000g = 7075 55s - 505g = 13
CalcQuot
(II): Calculate quotient
25s + 2000g = 7075 1s - 10g = 13
IdPropMult
(II): Identity Property of Multiplication
25s + 2000g = 7075 s - 10g = 13
AddEqn
(II): LHS+10g=RHS+10g
25s + 2000g = 7075 s = 10g + 13
Substitute
(I): s= 10g + 13
25( 10g + 13) + 2000g = 7075 s = 10g + 13
g = 3 s = 10g + 13
The vlogger bought 3 ounces of gold. It is possible to calculate the number of ounces of silver he purchased by substituting this value for s into Equation (II).
Therefore, the vlogger has 43 ounces of silver.
c To solve this system using the Elimination Method, it is important to remember that equivalent systems share the same solution. Equation (I) can be divided by 5 so that the variable s has the same coefficient in both equations.
25s + 2000g = 7075 & (I) 5s - 50g = 65 & (II)
▼
(I): Simplify
DivEqn
(I): .LHS /5.=.RHS /5.
25s + 2000g5 = 1415 5s - 50g = 65
WriteSumFrac
(I): Write as a sum of fractions
25s5 + 2000g5 = 1415 5s - 50g = 65
MovePartNumRight
(I): a* b/c=a/c* b
255s + 20005g = 1415 5s - 50g = 65
CalcQuot
(I): Calculate quotient
5s + 400g = 1415 5s - 50g = 65
Now, to continue solving using the Elimination Method, Equation (II) will be subtracted from Equation (I).
5s + 400g = 1415 & (I) 5s - 50g = 65 & (II)
SysEqnSub
(I): Subtract (II)
5s + 400g - ( 5s - 50g) = 1415 - 65 5s - 50g = 65
▼
(I): Solve for g
Distr
(I): Distribute - 1
5s + 400g - 5s + 50g = 1415 - 65 5s - 50g = 65
AddSubTerms
(I): Add and subtract terms
450g = 1350 5s - 50g = 65
DivEqn
(I): .LHS /450.=.RHS /450.
g = 3 5s - 50g = 65
The vlogger has 3 ounces of gold this time, too. This value can be substituted into Equation (II) to find out how many ounces of silver he has.
In this case, using the Elimination Method might be a little more intuitive than using the Substitution Method. Note that both of these methods are more reliable than solving the system graphically.
d To interpret the system in terms of the independence and consistency, these concepts should be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has exactly one solution, it is both consistent and independent.
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Example
Buying Equipment for the School Gym
As part of his student council duties, Tadeo is in charge of getting equipment for the school gym. At the moment, the gym needs new basketballs and volleyballs. Each basketball costs $75 and each volleyball costs $50. The gym's budget is $1400.
Tadeo can only go to the store once to buy the equipment. Each basketball weighs 22 ounces and each volleyball weighs 10 ounces. The total weight of the purchased balls is 364 ounces. For Tadeo to be able to spend all the budget in one go, the number of basketballs b and volleyballs v need to satisfy the following system of linear equations. 75b + 50v = 1400 & (I) 22b + 10v = 364 & (II)
a Solve the system using the Elimination Method.
b Verify the answer from Part A.
c Interpret the system in terms of the consistency and independence.
Answer
a See solution.
b See solution.
c The system is consistent and independent.
Hint
a What is the easiest way for a variable to have the same coefficient in both equations?
b Substitute the values from Part A into the original system of equations.
c A system of equations can be inconsistent, consistent independent, or consistent dependent.
Solution
a To solve the system using the Elimination Method, the coefficients of one variable have to be same to eliminate the term by adding or subtracting. Looking at the system, it can be noted that dividing Equation (I) by -5 will achieve this purpose.
75b + 50v = 1400 & (I) 22b + 10v = 364 & (II)
▼
(I): Simplify
DivEqn
(I): .LHS /(- 5).=.RHS /(- 5).
75b + 50v- 5 = - 280 22b + 10v = 364
WriteSumFrac
(I): Write as a sum of fractions
75b- 5 + 50v- 5 = - 280 22b + 10v = 364
MovePartNumRight
(I): a* b/c=a/c* b
75- 5b + 50- 5v = - 280 22b + 10v = 364
MoveNegDenomToFrac
(I): Put minus sign in front of fraction
- 755b + (- 505v) = - 280 22b + 10v = 364
AddNeg
a+(- b)=a-b
- 755b - 505v = - 280 22b + 10v = 364
CalcQuot
(I): Calculate quotient
- 15b -10v = - 280 22b + 10v = 364
Now, v has opposite coefficients. Therefore, by adding the equations, this variable will be eliminated.
- 15b -10v = - 280 & (I) 22b + 10v = 364 & (II)
SysEqnAdd
(I): Add (II)
- 15b -10v +( 22b+10v)= - 280+ 364 22b + 10v = 364
▼
(I): Solve for b
RemovePar
(I): Remove parentheses
- 15b -10v +22b+10v= - 280+364 22b + 10v = 364
AddSubTerms
(I): Add and subtract terms
7b= 84 22b + 10v = 364
DivEqn
(I): .LHS /7.=.RHS /7.
b= 12 22b + 10v = 364
Since b=12, Tadeo can buy and carry 12 basketballs. To find the number of volleyballs he can buy, this value will be substituted into Equation (II).
Therefore, Tadeo can buy and carry 12 basketballs and 10 volleyballs.
b To verify the solution, the values found in Part A need to be substituted into the system of equations.
| Equation (I) | Equation (II) | |
|---|---|---|
| Equation | 75b + 50v = 1400 | 22b + 10v = 364 |
| Substitute | 75( 12) + 50( 10) = 1400 | 22( 12) + 10( 10) = 364 |
| Simplify | 1400 = 1400 ✓ | 364 = 364 ✓ |
The values verify both equations of the system. Therefore, the solution is correct.
c To interpret the system in terms of independence and consistency, these concepts will be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has exactly one solution, the system is both consistent and independent.
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Example
Tools in a Toolbox
Ignacio is looking through his dad's toolbox to see if there are any duplicates that can be sold at a garage sale. He discovers that there are some extra hammers and wrenches.
In fact, the number of wrenches is equal to the difference between 13 and twice the number of hammers. He figures out that if he sells each hammer at $7 and each wrench at $3.50, he will make a profit of $45.50. If h and w are the number of hammers and wrenches, respectively, the information can be written as a system of equations. w = 13 - 2h & (I) 7h + 3.50w = 45.50 & (II)
a Solve the system of equations using the Elimination Method.
b Interpret the system in terms of consistency and independence.
Answer
a See solution.
b The system is consistent and dependent.
Hint
a In Equation (I), group the variable terms on the same side.
b A system of equations can be inconsistent, consistent independent, or consistent dependent.
Solution
a To solve the system using the Elimination Method, all the variables should be on the same side of the equation in both equations. This can be achieved by adding 2h to both sides of Equation (I).
w = 13 - 2h & (I) 7h + 3.50w = 45.50 & (II) ⇕ 2h+w = 13 & (I) 7h + 3.50w = 45.50 & (II) Now, a variable needs to have the same or opposite coefficients in both equations. This can be achieved by multiplying Equation (I) by 3.5.
2h+w = 13 & (I) 7h + 3.5w = 45.5 & (II)
MultEqn
(I): LHS * 3.5=RHS* 3.5
(2h+w)3.5 = 45.5 7h + 3.5w = 45.5
Distr
(I): Distribute 3.5
7h+3.5w = 45.5 7h + 3.5w = 45.5
SysEqnSub
(II): Subtract (I)
7h+3.5w = 45.5 7h + 3.5w - ( 7h + 3.5w) = 45.5 - 45.5
Distr
(II): Distribute - 1
7h+3.5w = 45.5 7h + 3.5w - 7h - 3.5w = 45.5 - 45.5
SubTerms
(II): Subtract terms
7h+3.5w = 45.5 0 = 0 ✓
By doing these operations, Equation (II) becomes an identity, or a true statement. This means that this system of equations has infinitely many solutions. Realizing this, Ignacio thinks that it is too big of a coincidence and decides to count the tools again and change the prices before going to the garage sale.
b To interpret the system in terms of consistency and independence, these concepts should be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has infinitely many solutions, the system is consistent and dependent. In a system of linear equations, if the system is dependent, then the equations that make the system are equations of coincidental lines.
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Example
The Lengths of Two Tracks
LaShay is on the track and field team at school. She has two tracks nearby her home where she trains for races. She is trying to remember the lengths of the tracks.
a Solve the system of equations using the Elimination Method.
b Interpret the system in terms of consistency and independence.
Answer
a See solution.
b The system is inconsistent.
Hint
a Make the coefficients of a variable term the same in both equations.
b A system of equations can be inconsistent, consistent independent, or consistent dependent.
Solution
a To solve the system using the Elimination Method, a variable term first needs to have the same coefficient in both equations. To achieve this, the first equation will be multiplied by 3.
2x + y = 5 & (I) 6x + 3y = 18 & (II)
6x + 3y = 15 6x + 3y = 18
The y-variable has the same coefficient in both equations. Now, Equation (II) can be subtracted from Equation (I).
6x + 3y = 15 & (I) 6x + 3y = 18 & (II)
SysEqnSub
(I): Subtract (II)
6x + 3y - ( 6x + 3y) = 15 - 18 6x + 3y = 18
0 = -3 * 6x + 3y = 18
Equation (I) was reduced to a false statement. Because of this, the equation has no solution. Knowing this, LaShay realizes that she does not remember the lengths of the tracks and decides to properly measure the tracks to prepare her training plan.
b To interpret the system in terms of consistency and independence, these concepts should be reviewed.
| Concept | Definition |
|---|---|
| Consistent System | A system of equations that has at least one solution. |
| Inconsistent System | A system of equations that has no solution. |
| Dependent System | A system of equations with infinitely many solutions. |
| Independent System | A system of equations that has exactly one solution. |
Since the system has no solution, it is an inconsistent system.
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Pop Quiz
Practice Solving Systems of Linear Equations
Solve the system of linear equations to find the values of x and y.
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Closure
Finding Out How Many Toys Are in Each Box
At the beginning of this lesson, a system of equations was introduced for the number of toys in a box. In both equations, x is the number of robots and y the number of cars. 45x + 110y = 20 & (I) 15x + 5y = 500 & (II) Solve this system without graphing. How many robots and cars are there in each box?
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Hint
Multiply or divide an equation by a number so that a variable has the same coefficient in both equations.
Solution
This system of linear equations can be solved without graphing using the Elimination Method. To use this method, Equation (I) will be multiplied by 10 to get rid of the fractions, while Equation (II) will be divided by 5 to make the coefficients of the y-variable equal.
45x + 110y = 20 & (I) 15x + 5y = 500 & (II)
▼
(I), (II): Simplify
MultEqn
(I): LHS * 10=RHS* 10
( 45x + 110y)10 = 200 15x + 5y = 500
Distr
(I): Distribute 10
45x(10) + 110y(10) = 200 15x + 5y = 500
CommutativePropMult
(I): Commutative Property of Multiplication
45(10)x + 110(10)y = 200 15x + 5y = 500
MoveRightFacToNum
(I): a/c* b = a* b/c
405x + 1010y = 200 15x + 5y = 500
CalcQuot
(I): Calculate quotient
8x + 1y = 200 15x + 5y = 500
DivEqn
(II): .LHS /5.=.RHS /5.
8x + 1y = 200 15x + 5y5 = 100
WriteSumFrac
(II): Write as a sum of fractions
8x + 1y = 200 15x5 + 5y5 = 100
MovePartNumRight
(II): a* b/c=a/c* b
8x + 1y = 200 155x + 55y = 100
CalcQuot
(II): Calculate quotient
8x + 1y = 200 3x + 1y = 100
(I), (II): Identity Property of Multiplication
8x + y = 200 3x + y = 100
Now that the variable y has the same coefficient in both equations, Equation (II) will be subtracted from Equation (I). Then, Equation (I) will be solved for the x-variable.
8x + y = 200 & (I) 3x + y = 100 & (II)
SysEqnSub
(I): Subtract (II)
8x + y-( 3x+y) = 200- 100 3x + y = 100
x = 20 3x + y = 100
Therefore, there are 20 robots in each box. Now, the value of x will be substituted into Equation (II) to find the value of y.
There are 40 cars in each box.
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Solving Systems of Equations by Elimination
Exercise 2.1
Dominika is running on a treadmill that can be set to different speeds. On the display she can read the amount of calories she burns during a workout session on the treadmill.
Dominika usually sets the treadmill to a speed of 8 kilometers per hour, which is her low speed. After a while, she usually increases the speed to 12 kilometers per hour, her high speed. The table shows two different workout sessions.
| Low speed | High speed | Calories Burnt | |
|---|---|---|---|
| Workout 1 | 20 min | 10 min | 300 |
| Workout 2 | 10 min | 15 min | 280 |
How many calories per minute does Dominika burn when she runs at high speed?
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We need to create two equations, one describing the first workout and the other describing the workout. Let x and y be the calories per minute that Dominika burns when she runs at low and high speed, respectively.
First Workout
Dominika runs 20 minutes at low speed and 10 minutes at high speed. With this information, we can write expressions for the calories she burns at these speeds. Low Speed: 20x calories High Speed: 10y calories The sum of these expressions is 300, which is the amount of calories she burns in the first session. 20x+10y=300
Second Workout
Similar to the first workout, we can write an equation describing the amount of calories Dominika burns when running at high and low speed during her second workout session. 10x+15y=280
System of Equations
We can combine these equations to create a system of equations, which we can solve by using the Elimination Method.
We can substitute 13 for y in Equation (I) to find the value of x. However, knowing that y=13 is enough information to state that Dominika burns 13 calories per minute when running at a higher speed.
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E
Hint & Answer
S
Solution
Consider the following system of equations. 2x+3y=5 & (I) 7x+5y=12 & (II) The first equation of an equivalent system is the same as the first equation of the given system. However, the second equation is different. In this equation, the coefficient of the x-variable is 13 and the constant is 27. Find the coefficient of the y-variable.
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An equivalent system has the same solution as the original system. Therefore, let's start by finding the solution to the original system.
Solving the Original System
To solve this system, we will use the Elimination Method.
Next, to find the value of x, we will substitute y=1 in Equation (I).
The solution to the original system is (1,1).
Finding an Equivalent System
We know that the coefficient of x is 13 and that the constant is 27. 13x+ay=27 The new and the original systems are equivalent. Therefore, (1,1) is also a solution to the new system. This means that the solution x=1, y=1, also satisfies the new equation. With this information in mind, we can find the value of a.
The coefficient of y is 14. 2x+3y=5 & (I) 13x+14y=27 & (II)
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Zosia swims every week in a river with a constant current. Against the current, she swims 150 meters in 200 seconds. With the current, she swims 100 meters in 80 seconds. Assuming that Zosia always swims at the same constant speed, what is the speed of the current?
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Speed is distance over time. speed=distance/time With the given information, we have enough to calculate Zosia's speed when she swims with and against the current. With Current:& s=100/80 [1em] Against Current:& s=150/200 When swimming with the current, Zosia's total speed is the sum of her swimming speed the speed of the current. Conversely, when she swims against the current, the speed of the current is subtracted from the her swimming speed to find the total speed. There are two unknown variables, the current's speed and Zosia's swimming speed. Let them be r_C and r_Z, respectively. With Current:& r_Z+r_C=100/80 [1em] Against Current:& r_Z-r_C=150/200 Now we have two equations that relate the speeds of Zosia and the current. If we combine them, we can write a system of equations which we can solve by using the Elimination Method.
The speed of the current is 0.25 meters per second.
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Solving Systems of Equations by Elimination
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