7.4: Solving Systems of Equations by Multiplication
Learning objectives
 Solve a linear system by multiplying one equation.
 Solve a linear system of equations by multiplying both equations.
 Compare methods for solving linear systems.
 Solve realworld problems using linear systems by any method.
Introduction
We have now learned three techniques for solving systems of equations.
 Graphing, Substitution and Elimination (through addition and subtraction).
Each one of these methods has both strengths and weaknesses.
 Graphing is a good technique for seeing what the equations are doing, and when one service is less expensive than another, but graphing alone to find a solution can be imprecise and may not be good enough when an exact numerical solution is needed.
 Substitution is a good technique when one of the coefficients in an equation is +1 or 1, but can lead to more complicated formulas when there are no unity coefficients.
 Addition or Subtraction is ideal when the coefficients of either \begin{align*}x\end{align*} or \begin{align*}y\end{align*} match in both equations, but so far we have not been able to use it when coefficients do not match.
In this lesson, we will again look at the method of elimination that we learned in Lesson 7.3. However, the equations we will be working with will be more complicated and one can not simply add or subtract to eliminate one variable. Instead, we will first have to multiply equations to ensure that the coefficients of one of the variables are matched in the two equations.
Quick Review: Multiplying Equations
Consider the following questions
 If 10 apples cost $5, how much would 30 apples cost?
 If 3 bananas plus 2 carrots cost $4, how much would 6 bananas plus 4 carrots cost?
You can look at the first equation, and it should be obvious that each apple costs $0.50. 30 apples should cost $15.00.
Looking at the second equation, it is not clear what the individual price is for either bananas or carrots. Yet we know that the answer to question two is $8.00. How?
If we look again at question one, we see that we can write the equation \begin{align*}10a = 5 \end{align*} (\begin{align*}a\end{align*} being the cost of one apple).
To find the cost of 30, we can either solve for a them multiply by 30, or we can multiply both sides of the equation by three.
\begin{align*}30a & = 15 \\ a & = \frac{1}{2}\ \text{or}\ 0.5\end{align*}
Now look at the second question. We could write the equation \begin{align*}3b + 2c = 4\end{align*}.
We see that we need to solve for \begin{align*}(6b + 4c)\end{align*} which is simply two times the quantity \begin{align*}(3b + 2c)\end{align*}!
Algebraically, we are simply multiplying the entire equation by two.
\begin{align*}2(3b + 2c) & = 2\cdot 4 & \text{Distribute and multiply.}\\ 6b + 4c & = 8\end{align*}
So when we multiply an equation, all we are doing is multiplying every term in the equation by a fixed amount.
Solving a Linear System by Multiplying One Equation
We can multiply every term in an equation by a fixed number (a scalar), it is clear that we could use the addition method on a whole new set of linear systems. We can manipulate the equations in a system to ensure that the coefficients of one of the variables match. In the simplest case, the coefficient as a variable in one equation will be a multiple of the coefficient in the other equation.
Example 1
Solve the system.
\begin{align*}7x + 4y & = 17\\ 5x 2y & = 11\end{align*}
It is quite simple to see that by multiplying the second equation by two the coefficients of \begin{align*}y\end{align*} will be +4 and 4, allowing us to complete the solution by addition.
Take two times equation two and add it to equation one. Then divide both sides by 17 to find \begin{align*}x\end{align*}.
\begin{align*} & \quad \ \ 10x  4y = 22\\ & \underline{\ \ + (7x + 4y) = 17\;\;}\\ & \qquad \qquad 17y = 34\\ & \qquad \qquad \quad x = 2\end{align*}
Now simply substitute this value for \begin{align*}x\end{align*} back into equation one.
\begin{align*}7 \cdot 2 + 4y & = 17 && \text{Since}\ 7 \times 2 = 14,\ \text{subtract}\ 14\ \text{from both sides.}\\ 4y & = 3 && \text{Divide by}\ 4.\\ y & = 0.75\end{align*}
Example 2
Anne is rowing her boat along a river. Rowing downstream, it takes her two minutes to cover 400 yards. Rowing upstream, it takes her eight minutes to travel the same 400 yards. If she was rowing equally hard in both directions, calculate, in yards per minute, the speed of the river and the speed Anne would travel in calm water.
Step One First we convert our problem into equations. We need to know that distance traveled is equal to speed \begin{align*}x\end{align*} time. We have two unknowns, so we will call the speed of the river \begin{align*}x\end{align*}, and the speed that Anne rows at \begin{align*}y\end{align*}. When traveling downstream her total speed is the boat speed plus the speed of the river \begin{align*}(x + y)\end{align*}. Upstream her speed is hindered by the speed of the river. Her speed upstream is \begin{align*}(xy)\end{align*}.
\begin{align*}\text{Downstream Equation} && 2(x+y) & = 400 \\ \text{Upstream Equation} && 8(xy) & = 400\end{align*}
Distributing gives us the following system.
\begin{align*}2x + 2y & = 400 \\ 8x  8y & = 400\end{align*}
Right now, we cannot use the method of elimination as none of the coefficients match. But, if we were to multiply the top equation by four, then the coefficients of \begin{align*}y\end{align*} would be +8 and 8. Let’s do that.
\begin{align*}& \qquad 8x  8y = 1,600\\ & \underline{+ \ (8x  8y) = 400\;\;\;\;}\\ & \qquad \quad \ \ 16x = 2,000\end{align*}
Now we divide by 16 to obtain \begin{align*}x = 125\end{align*}.
Substitute this value back into the first equation.
\begin{align*}2(125 + y) & = 400 && \text{Divide both sides by}\ 2.\\ 125 + y & = 200 && \text{Subtract}\ 125\ \text{from both sides.}\\ y & = 75\end{align*}
Solution
Anne rows at 125 yards per minute, and the river flows at 75 yards per minute.
Solve a Linear System by Multiplying Both Equations
It is a straightforward jump to see what would happen if we have no matching coefficients and no coefficients that are simple multiples of others. Just think about the following fraction sum.
\begin{align*} \frac {1}{2} + \frac {1}{3} = \frac {3}{6} + \frac {2}{6} = \frac {5}{6}\end{align*}
This is an example of finding a lowest common denominator. In a similar way, we can always find a lowest common multiple of two numbers (the lowest common multiple of 2 and 3 is 6). This way we can always find a way to multiply equations such that two coefficients match.
Example 3
Andrew and Anne both use the IHaul truck rental company to move their belongings from home to the dorm rooms on the University of Chicago campus. IHaul has a charge per day and an additional charge per mile. Andrew travels from San Diego, California, a distance of 2,060 miles in five days. Anne travels 880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays $1845, what does IHaul charge
a) per day?
b) per mile traveled?
First, we will setup our equations. Again we have two unknowns, the daily rate (we will call this \begin{align*}x\end{align*}), and the rate per mile (let’s call this \begin{align*}y\end{align*}).
\begin{align*}\text{Anne’s equation} && 3x + 880y & = 840 \\ \text{Andrew’s Equation} && 5x + 2060y & = 1845\end{align*}
We cannot simply multiply a single equation by an integer number in order to arrive at matching coefficients. But if we look at the coefficients of \begin{align*}x\end{align*} (as they are easier to deal with than the coefficients of \begin{align*}y\end{align*}) we see that they both have a common multiple of 15 (in fact 15 is the lowest common multiple). So this time we need to multiply both equations:
Multiply Anne's equation by five:
\begin{align*}15x + 4400y = 4200\end{align*}
Multiply Andrew's equation by three:
\begin{align*}15x + 6180y = 5535\end{align*}
Subtract:
\begin{align*}& \qquad 15x + 4400y = 4200\\ & \underline{ \ (15x + 6180y) = 5535\;\;\;\;}\\ & \qquad \quad \ \ 1780y = 1335\end{align*}
Divide both sides by 1780
\begin{align*}y=0.75\end{align*}
Substitute this back into Anne's equation.
\begin{align*}3x + 880(0.75) & = 840 && \text{Since}\ 880 \times 0.75 = 660,\ \text{subtract}\ 660\ \text{from both sides.}\\ 3x & = 180 && \text{Divide both sides by}\ 3.\\ x & =60\end{align*}
Solution
IHaul charges $60 per day plus $0.75 per mile.
Comparing Methods for Solving Linear Systems
Now that we have covered the major methods for solving linear equations, let’s review. For simplicity, we will look at the four methods (we will consider addition and subtraction one method) in table form. This should help you decide which method would be better for a given situation.
Method:  Best used when you...  Advantages:  Comment: 

Graphing  ...don’t need an accurate answer.  Often easier to see number and quality of intersections on a graph. With a graphing calculator, it can be the fastest method since you don’t have to do any computation.  Can lead to imprecise answers with noninteger solutions. 
Substitution  ...have an explicit equation for one variable (e.g. \begin{align*}y = 14x +2\end{align*})  Works on all systems. Reduces the system to one variable, making it easier to solve.  You are not often given explicit functions in systems problems, thus it can lead to more complicated formulas 
Elimination by Addition or Subtraction  ...have matching coefficients for one variable in both equations.  Easy to combine equations to eliminate one variable. Quick to solve.  It is not very likely that a given system will have matching coefficients. 
Elimination by Multiplication and then Addition and Subtraction  ...do not have any variables defined explicitly or any matching coefficients.  Works on all systems. Makes it possible to combine equations to eliminate one variable.  Often more algebraic manipulation is needed to prepare the equations. 
The table above is only a guide. You may like to use the graphical method for every system in order to better understand what is happening, or you may prefer to use the multiplication method even when a substitution would work equally well.
Example 4
Two angles are complementary when the sum of their angles is \begin{align*}90^{\circ}\end{align*}. Angles \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are complementary angles, and twice the measure of angle \begin{align*}A\end{align*} is \begin{align*}9^{\circ}\end{align*} more than three times the measure of angle \begin{align*}B\end{align*}. Find the measure of each angle.
First, we write out our two equations. We will use \begin{align*}x\end{align*} to be the measure of Angle \begin{align*}A\end{align*} and \begin{align*}y\end{align*} to be the measure of Angle \begin{align*}B\end{align*}. We get the following system
\begin{align*}x + y & = 90 \\ 2x & = 3y + 9\end{align*}
The first method we will use to solve this system is the graphical method. For this we need to convert the two equations to \begin{align*}y = mx +b\end{align*} form
\begin{align*}& x + y = 90 && \Rightarrow && y = x + 90 \\ & 2x = 3y + 9 && \Rightarrow && y = \frac {2}{3}x  3\end{align*}
The first line has a slope of 1 and a \begin{align*}y\end{align*}intercept of 90.
The second has a slope of \begin{align*}\frac{2}{3}\end{align*} and a \begin{align*}y\end{align*}intercept of 3.
In the graph, it appears that the lines cross at around \begin{align*}x = 55, y =35\end{align*} but it is difficult to tell exactly! Graphing by hand is not the best method if you need to know the answer with more accuracy!
Next, we will try to solve by substitution. Let’s look again at the system:
\begin{align*}x + y & = 90 \\ 2x & = 3y + 9\end{align*}
We have already seen that we can solve for \begin{align*}y\end{align*} with either equation in trying to solve the system graphically.
Solve the first equation for \begin{align*}y\end{align*}.
\begin{align*}y = 90 x\end{align*}
Substitute into the second equation
\begin{align*}2x & = 3(90 x) + 9 && \text{Distribute the}\ 3.\\ 2x & = 270 3x + 9 && \text{Add}\ 3x\ \text{to both sides.}\\ 5x & = 270 + 9 = 279 && \text{Divide by}\ 5.\\ x & = 55.8^{\circ}\end{align*}
Substitute back into our expression for \begin{align*}y\end{align*}.
\begin{align*}y = 90  55.8 = 34.2^{\circ}\end{align*}
Solution
Angle \begin{align*}A\end{align*} measures \begin{align*}55.8^{\circ}\end{align*}. Angle \begin{align*}B\end{align*} measures \begin{align*}34.2^{\circ}\end{align*}
Finally, we will examine the method of elimination by multiplication.
Rearrange equation one to standard form
\begin{align*} x + y = 90 \Rightarrow 2x + 2y = 180\end{align*}
Multiply equation two by two.
\begin{align*} 2x = 3y + 9 \Rightarrow 2x  3y = 9\end{align*}
Subtract.
\begin{align*}& \qquad 2x + 2y = 180\\ & \underline{ \ (2x  2y) = 9\;\;}\\ & \qquad \qquad \ 5y = 171\end{align*}
Divide both sides by 5
\begin{align*}y=34.2\end{align*}
Substitute this value into the very first equation.
\begin{align*}x+ 34.2 & = 90 && \text{Subtract}\ 34.2\ \text{from both sides.}\\ x & = 55.8^\circ\end{align*}
Solution
Angle \begin{align*}A\end{align*} measures \begin{align*}55.8^{\circ}\end{align*}. Angle \begin{align*}B\end{align*} measures \begin{align*}34.2^{\circ}\end{align*}.
Even though this system looked ideal for substitution, the method of multiplication worked well, also. Once the algebraic manipulation was performed on the equations, it was a quick solution. You will need to decide yourself which method to use in each case you see from now on. Try to master all techniques, and recognize which technique will be most efficient for each system you are asked to solve.
Multimedia Link For even more practice, we have this video. One common type of problem involving systems of equations (especially on standardized tests) is “age problems”. In the following video the narrator shows two examples of age problems, one involving a single person and one involving two people. Khan Academy Age Problems (7:13)
.
Review Questions
 Solve the following systems using multiplication.
 \begin{align*}5x10y = 15\!\\ 3x 2y = 3\end{align*}
 \begin{align*}5x  y = 10\!\\ 3x 2y = 1\end{align*}
 \begin{align*}5x + 7y = 15\!\\ 7x  3y = 5\end{align*}
 \begin{align*}9x + 5y = 9\!\\ 12x + 8y = 12.8\end{align*}
 \begin{align*}4x  3y = 1\!\\ 3x  4y = 4\end{align*}
 \begin{align*}7x3y = 3\!\\ 6x + 4y =3\end{align*}
 Solve the following systems using any method.
 \begin{align*}x= 3y\!\\ x2y = 3\end{align*}
 \begin{align*}y = 3x + 2\!\\ y = 2x + 7\end{align*}
 \begin{align*}5x5y = 5\!\\ 5x + 5y = 35\end{align*}
 \begin{align*}y = 3x3\!\\ 3x2y + 12 = 0\end{align*}
 \begin{align*}3x4y = 3\!\\ 4y + 5x = 10\end{align*}
 \begin{align*}9x2y = 4\!\\ 2x 6y = 1\end{align*}
 Supplementary angles are two angles whose sum is \begin{align*}180^{\circ}\end{align*}. Angles \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are supplementary angles. The measure of Angle \begin{align*}A\end{align*} is \begin{align*}18^{\circ}\end{align*} less than twice the measure of Angle \begin{align*}B\end{align*}. Find the measure of each angle.
 A farmer has fertilizer in 5% and 15% solutions. How much of each type should he mix to obtain 100 liters of fertilizer in a 12% solution?
 A 150 yards pipe is cut to provide drainage for two fields. If the length of one piece is three yards less that twice the length of the second piece, what are the lengths of the two pieces?
 Mr. Stein invested a total of $100, 000 in two companies for a year. Company A’s stock showed a 13% annual gain, while Company B showed a 3% loss for the year. Mr. Stein made an 8% return on his investment over the year. How much money did he invest in each company?
 A baker sells plain cakes for $7 or decorated cakes for $11. On a busy Saturday the baker started with 120 cakes, and sold all but three. His takings for the day were $991. How many plain cakes did he sell that day, and how many were decorated before they were sold?
 Twice John’s age plus five times Claire’s age is 204. Nine times John’s age minus three times Claire’s age is also 204. How old are John and Claire?
Review Answers

 \begin{align*}x=0, y = 1.5\end{align*}
 \begin{align*}x=3, y=5\end{align*}
 \begin{align*}x=1.25, y=1.25\end{align*}
 \begin{align*}x=\frac{2}{3}, y=\frac{3}{5}\end{align*}
 \begin{align*}x=\frac{8}{7}, y=\frac{13}{7}\end{align*}
 \begin{align*}x=\frac{3}{46}, y=\frac{39}{46}\end{align*}
 \begin{align*}x = 9, y= 3\end{align*}
 \begin{align*}x=1, y=5\end{align*}
 \begin{align*}x=4, y=3\end{align*}
 \begin{align*}x=2, y=3\end{align*}
 \begin{align*}x=\frac{13}{8}, y=\frac{15}{32}\end{align*}
 \begin{align*}x=\frac{13}{25}, y=\frac{17}{50}\end{align*}
 \begin{align*}A = 114^{\circ}, B = 66^{\circ}\end{align*}
 30 liters of 5%, 70 liters of 15%
 51 yards and 99 yards
 $68, 750 in Company A, $31, 250 in Company B
 74 plain, 43 decorated
 John is 32, Claire is 28
Notes/Highlights Having trouble? Report an issue.
Color  Highlighted Text  Notes  

Show More 