# 7.6: Comparing Methods for Solving Linear Systems

**At Grade**Created by: CK-12

**Practice**Comparing Methods for Solving Linear Systems

### Comparing Methods for Solving Linear Systems

Now that we’ve covered the major methods for solving linear equations, let’s review them. For simplicity, we’ll look at them in table form. This should help you decide which method would be best 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 non-integer 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, so you may have to do extra work to get one of the equations into that form. |

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 might prefer to use the graphical method for every system in order to better understand what is happening, or you might prefer to use the multiplication method even when a substitution would work just as well.

#### Measuring Angles

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 2 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*}

First, we’ll solve this system with 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 \qquad \ \Rightarrow y = -x + 90\\ & 2x = 3y + 9 \qquad \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, and the second line has a slope of \begin{align*}\frac{2}{3}\end{align*} and a \begin{align*}y-\end{align*}intercept of -3. The graph looks like this:

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 in this case!

#### Solving by Substituion

In this example, we’ll try solving by substitution. Let’s look again at the system:

\begin{align*}x + y &= 90\\ 2x &= 3y + 9\end{align*}

We’ve already seen that we can start by solving either equation for \begin{align*}y\end{align*}, so let’s start with the first one:

\begin{align*}y = 90 - x\end{align*}

Substitute into the second equation:

\begin{align*}& 2x = 3(90 - x) + 9 && distribute \ the \ 3:\\ & 2x = 270 - 3x + 9 && add \ 3x \ to \ both \ sides:\\ & 5x = 270 + 9 = 279 && 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*}

**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*}.

#### Solving by Elimination

Finally, in this example, we’ll try solving by elimination (with multiplication):

Rearrange equation one to standard form:

\begin{align*}& x + y = 90 \qquad \Rightarrow 2x + 2y = 180\end{align*}

Multiply equation two by 2:

\begin{align*}&2x = 3y + 9 \qquad \Rightarrow 2x - 3y = 9\end{align*}

Subtract:

\begin{align*}& \quad \qquad \qquad \qquad 2x + 2y = 180\\ & \qquad \qquad \ - \ \ (2x - 3y) = -9\\ & \qquad \qquad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \qquad \qquad \qquad \qquad 5y = 171\\ \\ & \text{Divide by 5 to obtain} \ y = 34.2^\circ\end{align*}

Substitute this value into the very first equation:

\begin{align*}x + 34.2 &= 90 && subtract \ 34.2 \ from \ both \ sides:\\ x &= 55.8^\circ\end{align*}

**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 too. Once the equations were rearranged properly, the solution was quick to find. You’ll need to decide yourself which method to use in each case you see from now on. Try to master all the techniques, and recognize which one will be most efficient for each system you are asked to solve.

### Example

#### Example 1

Solve the system \begin{align*}\begin{cases} 5s+2t=6\\ 9s+2t=22\end{cases}\end{align*}.

Since these equations are both written in standard form, and both have the term \begin{align*}2t\end{align*} in them, we will use elimination by subtracting. This will cause the \begin{align*}t\end{align*} terms to cancel out and we will be left with one variable, \begin{align*}s\end{align*}, which we can then isolate.

\begin{align*}& \qquad \ 5s+2t=6\\ &\underline{\;\; - \ (9s+2t = 22) \;\;}\\ & \qquad \ -4s+0t =-16\\ & \qquad \ -4s=-16\\ & \qquad \ s=4\end{align*}

\begin{align*}5(4)+2t&=6\\ 20+2t&=6\\ 2t&=-14\\ t&=-7\end{align*}

The solution is \begin{align*}(4,-7)\end{align*}.

### Review

Solve the following systems using any method.

- \begin{align*}x = 3y\!\\ x - 2y = -3\end{align*}
- \begin{align*}y = 3x + 2\!\\ y = -2x + 7\end{align*}
- \begin{align*}5x - 5y = 5\!\\ 5x + 5y = 35\end{align*}
- \begin{align*}y = -3x - 3\!\\ 3x - 2y + 12 = 0\end{align*}
- \begin{align*}3x - 4y = 3\!\\ 4y + 5x = 10\end{align*}
- \begin{align*}9x - 2y = -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-yard 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?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 7.6.

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |

### Image Attributions

Here you'll learn how to decide which method of solving a system of linear equations is the best one for the situation.