In this section we'll focus on finding the point (or sometimes points) where two lines intersect. What we're after is an ordered pair of coordinates, $(x, y).$ That means we'll be trying to solve for two variables, x and y. We rarely get something for nothing in mathematics, and that includes this case. In order to find the values of these two variables, we'll need two pieces of information, namely the equations of the two lines.

Take a minute to remember how lines work. A fine example of a linear equation is

$$2y + 3x = 4$$

This equation defines a set of points, (x, y) which satisfy this equation (make the equal sign true). No points (x, y) not on the line will satisfy the equation. Here is a list of points lying on and off of our line. Check them using the formula to confirm:

On line |
Not on line |

(-2, 5) | (-1, 3) |

(0, 2) | (1, 1) |

(2, -1) | (3, 4) |

Now when two lines intersect on a graph, they *share* just that one point. It's a special point, because it satisfies the equations of both lines. That's often a point of interest for us, and this section will cover ways to find it using examples.

If two non-identical lines in a plane intersect, then there is one, and only one point, $(x, y)$ in that plane that satisfies the equations of both lines.

Find the point of intersection of lines $x + y = 2$ and $y = 1.$

**Solution**

$$ \begin{align} x + y &= 2 \\[3pt] y &= 1 \end{align}$$

Both lines are plotted below. If we rearrange the first to $y = mx + b$ form, we get $y = -x + 2,$ a line that passes through (0, 2) and has a slope of -1. The second is the horizontal line passing through (0, 1).

Now for each of these examples, we'll find the point of intersection three different ways, by substitution, by elimination, and by the transitive property. Which one you use will be entirely up to you, but some methods easier and more direct than others in certain circumstances.

The substitution method involves solving one of our two equation in terms of one variable, then plugging the expression for that variable into its appropriate place in the other equation. Let's use it on the current example.

In this system of two equations, we already have $y = 1$, so that's a good target for substitution. Now substitute 1 for y in the other equation, $x + y = 2$ to get

$$ \begin{align} x + y &= 2 \; \longleftarrow \; y = 1 \\[4pt] x + 1 &= 2 \\[4pt] x &= 1 \end{align}$$

Now that we have our x value, we use either equation to solve for y. It's obvious from the second that $y = 1$, and our point of intersection is **(1, 1)**, as shown in the graph.

When we solve systems of equations by elimination, we are relying on the fact that the sides of an equation – across the equal sign – are equivalent. Therefore if we add or subtract two equations (left side ± left side, right side ± right side, the result should still be a valid equation.

Now consider our system of equations

$$ \begin{align} x + y &= 2 \\[4pt] y &= 1 \end{align}$$

Now let's subtract the lower equation from the upper (nothing wrong with subtracting the upper from the lower, but one way generally gets you to a solution with less trouble):

$$ \begin{matrix} \phantom{000} x + y = 2 \\[1pt] \underline{\phantom{0000} -(y = 1)} \\[1pt] \phantom{000000} x = 1 \end{matrix}$$

... and that once again leads us to y = 1, for an intersection point of **(1, 1)**.

Finally, the transitive property tells us that if two things are equal to the same thing, then they are equal to each other. If we look at our two equations in this light:

$$ \begin{align} y &= 2 - x \\[3pt] y &= 1 \end{align}$$

If this is true, then it must also be true that $2 - x = 1,$ or $x = 1,$ which leads us once again to our intersection point. This method isn't very different from elimination, but sometimes it's easier to wrap your brain around. We'll continue employing all three methods in the next examples.

Find the point of intersection of lines $x + 3y = 1$ and $2y - x = -1.$

**Solution**

$$ \begin{align} x + 3y &= 1 \\[2pt] -x + 2y &= -1 \end{align}$$

If we rearrange these to $y = mx + b$ form, we get:

$$ \begin{align} y &= \frac{-1}{3} x + \frac{1}{3} \\[2pt] y &= \frac{1}{2} x - \frac{1}{2} \end{align}$$

The functions are plotted in the graph below. In this form, they're set up perfectly to use the **transitive property**. To find x, we'd solve

$$\frac{-1}{3} x + \frac{1}{3} = \frac{1}{2} x - \frac{1}{2}$$

Well, this isn't impossible, but it is a little messy because of all the fractions. Let's pass on the transitive property method for a better way.*

If we solve the second equation for y, we get $y = \frac{1}{2} x - \frac{1}{2}.$ If we plug that value of y in for the y in the first equation, we get:

$$ \begin{align} 3\left( \frac{1}{2} x - \frac{1}{2} \right) &= -x + 1 \\[5pt] \frac{3}{2} x - \frac{3}{2} &= -x + 1 \\[5pt] \frac{5}{2} x &= \frac{5}{2} \\[5pt] x &= 1 \end{align}$$

Then the y value can be found from either one of the other two equations. The second gives

$$ \begin{align} 2y &= 1 - 1 \\[2pt] y &= 0 \end{align}$$

So our point of intersection is **(1, 0)**, which goes along with what we graphed.

Elimination turns out to be the method of choice for these equations. Look at them lined up and notice that if they were added, the x's would subtract to zero:

$$ \begin{matrix} 3y = -x + 1 \\[2pt] \underline{2y = \; x - 1} \\[2pt] 5y = 0\phantom{0000} \end{matrix}$$

... and x follows from y. Here you see that there's often one method that leads most easily to the solution. Elimination is it for this one.

*

Find the point of intersection of lines $y = -x + 3$ and $y = 4x - 2.$

**Solution**

$$ \begin{align} y &= -x + 3 \\[2pt] y &= 4x - 2 \end{align}$$

They are plotted below:

These are perfectly set up to use the transitive property. If $y = -x + 3$ and $y = 4x - 2,$ then

$$ \begin{align} -x + 3 &= 4x - 2 \\[2pt] 5x &= 5 \\[2pt] x &= 1 \end{align}$$

Then y is easy to find with either equation:

$$ \begin{align} y &= -x + 3 \; \leftarrow \text{ first equation} \\[2pt] y &= -1 + 3 \\[2pt] y &= 2 \end{align}$$

So our point of intersection is **(1, 2)**.

I'll leave it to you do do this one by elimination (subtract the equations to eliminate y - same result as what we did here) or substitution. You should get the same result.

There are two exceptions to the case in which planar lines intersect at one point. They either

- Never intersect, in which case they're
**parallel**, or - They intersect at every point (an infinite number of points) because they are
**coincident**(the same line).

Recall that **parallel lines** have the same slope but different positions in the plane (usually measured by the y-intercept). And two linear equations that have the same slope and intercept are coincident, or not **linearly-independent**. If two lines are not linearly independent, then one can be obtained from the other by multiplying it by a constant. Let's do an algebraic example of each so you can see what happens when you encounter these cases.

Consider the lines $x - 2y = 7$ and $y = \frac{1}{2} x + \frac{7}{2}.$ Do they intersect? If so, where?

**Solution**

$$ \begin{align} x - 2y &= 7 \\[2pt] y &= \frac{1}{2} x + \frac{7}{2} \end{align}$$

Let's substitute $x = 2y + 7$ (a rearrangement of the first equation) into the second equation:

$$ \begin{align} y &= \frac{1}{2} (2y + 7) + \frac{7}{2} \\[2pt] y &= y - \frac{7}{2} + \frac{7}{2} \\[2pt] y &= y \end{align}$$

This result is true for any x at all, so these "two" lines intersect at every point in their (infinite) domains. They are the same line. The second is just the first multiplied by 2, rearranged and divided by 4.

In mathematics, the word coincident refers to figures in a plane that share all of their points. This can include identical lines, circles with the same radius and center point, parabolas with the same equation, and so on.

Parallel lines on a plane are those with the same slope, but a different location – as specified by the y-intercept, for example. Consider these two lines:

$$ \begin{align} y &= 2x - 4 \\[2pt] y &= 2x + 7 \end{align}$$

If we solve these by the transitive property (or elimination of y), we get

$$ \begin{align} 2x - 4 &= 2x + 7 \\[2pt] -4 &= 7 \end{align}$$

This result is *never* true. It's not true for any x in the domain of either line, so the two have no points in common. They are parallel. This system of equations has no solution.

When two lines are **parallel**, the solution to the system will be a __contradiction__, like **2 = 3**. It means that there is no value of x in the domain of either line (both -∞ to ∞) that will satisfy both equations.

When two lines are **coincident**, the solution to the system will be a statement that is __always true__, regardless of x, such as **5 = 5**. It means that every point on one line is also on the other.

Solve these systems of equations using the most convenient method for you. One or two ways will be used in the solutions.

1. |
$$ \begin{align} 2x + 3y &= 7 \\ x &= 2 \end{align}$$ ## SolutionUse substitution of x = 2 into the first equation to find y: $$ \begin{align} 2(2) + 37 &= 7 \\ 4 + 3y &= 7 \\ 3y &= 3 \\ y &= 1 \end{align}$$ The point of intersection is |

2. |
$$ \begin{align} x - 2y &= 4 \\ 3x + y &= 5 \end{align}$$ ## SolutionSolve the first equation for x, then substitute that into the second: $$ \begin{align} x &= 2y + 4 \\ 3(2y + 4) + y &= 5 \\ 6y + 12 + y &= 5 \\ 7y &= -7 \\ y &= -1 \end{align}$$ Now use either equation to find x: $$ \begin{align} x - 2y &= 4 \rightarrow x = 2y + 4 \\ x &= 2(-1) + 4 \\ x &= 2 \end{align}$$ The intersection point is |

3. |
$$ \begin{align} 3x + 4y &= 12 \\ -3x - y &= -4 \end{align}$$ ## SolutionThese equations are set up nicely for elimination of x through addition: $$ \begin{matrix} 3x + 4y = 12 && \\ \underline{-3x - y = -4} && \leftarrow \text{ add}\\ \phantom{0000}3y = 8 && \\[5pt] \phantom{00000}y = \frac{8}{3} \end{matrix}$$ Now use either equation to find x: $$ \begin{align} 3x + 4\left( \frac{8}{3} \right) &= 12 \\ 3x + \frac{32}{3} &= \frac{36}{3} \\ 3x &= \frac{4}{3} \\ x &= \frac{4}{9} \end{align}$$ The point of intersection is $\left( \frac{4}{9}, \frac{8}{3} \right)$ |

4. |
$$ \begin{align} x - y &= 7 \\ 7x - y &= 2 \end{align}$$ ## SolutionSubtracting these equations (either way) will eliminate y: $$ \begin{matrix} x - y = 7 && \\ \underline{7x - y = 2} && \leftarrow \text{ subtract}\\ \phantom{0}-6x = 5 && \\ \phantom{00000}x = \frac{-5}{6} && \end{matrix}$$ Now use the first equation to get y: $$ \begin{align} y &= x - 7 \\ y &= -\frac{5}{6} - \frac{42}{6} \\[3pt] y &= \frac{-47}{6} \end{align}$$ The point of intersection is $\left( \frac{-5}{6}, \frac{-47}{6} \right)$ |

5. |
$$ \begin{align} -x - 3y &= 11 \\ 2x + 5y &= -8 \end{align}$$ ## SolutionWe'll use elimination. Multiply the first equation by 2, then add them. $$ \begin{matrix} -2x - 6y = 22 && \\ \underline{\phantom{0}2x + 5y = -8} && \leftarrow \text{ add}\\ \phantom{000000}y = 14 && \end{matrix}$$ Now use one of the other equations (either will work; choose the simplest one) to find x. $$ \begin{align} -x - 3y &= 11 \\ x &= -3y - 11 \\ x &= -3(14) - 11 \\ x &= -42 - 11 = -53 \end{align}$$ The point of intersection is |

6. |
$$ \begin{align} y &= 3x - 11 \\ y &= 2x + 1 \end{align}$$ ## SolutionThis system is perfectly set up to eliminate y using the transitive property: $$ \begin{align} 3x - 11 &= 2x + 1 \\ x &= 12 \end{align}$$ Now use either equation to find y from x: $$ \begin{align} y &= 3x - 11 \\ y &= 3(12) - 11 \\ y &= 36 - 11 = 25 \end{align}$$ The point of intersection is |

7. |
$$ \begin{align} x - 2y &= 8 \\ 2y &= x + 22 \end{align}$$ ## SolutionRearrange and subtract these two equations: $$ \begin{matrix} x - 2y = 8 && \\ \underline{\phantom{00}x - 2y = -22} && \leftarrow \text{ subtract}\\ \phantom{00000}0 = 30 && \end{matrix}$$ This result is never true for any x in the domain of either line, therefore the system has |

8. |
$$ \begin{align} \frac{1}{2}x - \frac{2}{3}y &= 4 \\ x - y &= \frac{7}{2} \end{align}$$ ## SolutionMultiply the first equation by -2, then add the equations to eliminate x: $$ \begin{matrix} x - \frac{4}{3}y = 8 && \\ \underline{\phantom{00}x - y = \frac{7}{2}} && \leftarrow \text{ add}\\ \phantom{00}-\frac{1}{3}y = \frac{9}{2} && y = \frac{-27}{2} \end{matrix}$$ Now use the second equation (it's simpler) to find x from y: $$ \begin{align} x &= \frac{7}{2} + y \\ x &= \frac{7}{2} - \frac{27}{2} \\ x &= \frac{-20}{2} = -10 \end{align}$$ The point of intersection is $(-10, -\frac{27}{2}).$ |

9. |
$$ \begin{align} x + 2y &= 7 \\ y &= -\frac{1}{2}x + \frac{7}{2} \end{align}$$ ## SolutionThis system is set up for easy substitution of y from the second equation into the first: $$ \begin{align} x + 2 \left( -\frac{1}{2}x + \frac{7}{2} \right) &= 7 \\ x - x + 7 &= 7 \\ 7 &= 7 \end{align}$$ This result is always true for any x in the domain of either line, so these lines intersect at every point; they are |

10. |
$$ \begin{align} x + \frac{1}{2}y &= 5 \\ 2x - 3y &= -4 \end{align}$$ ## SolutionMultiply the first equation by 2, then subtract the two $$ \begin{matrix} 2x + y = 10 && \\ \underline{2x - 3y = -4} && \leftarrow \text{ subtract}\\ \phantom{000}4y = 14 && y = \frac{7}{2} \end{matrix}$$ Now substitute y into the first equation to find x: $$ \begin{align} x &= -\frac{1}{2} y + 5 \\ x &= -\frac{1}{2}\left( \frac{7}{2} \right) + 5 \\ x &= -\frac{7}{4} + \frac{20}{4} = \frac{13}{4} \end{align}$$ The point of intersection is $(\frac{13}{4}, \frac{7}{2}).$ |

11. |
$$ \begin{align} x + y &= 2 \\ x - y &= 2 \end{align}$$ ## SolutionIf we add these two equations just as they are, we get $$ \begin{align} 2x &= 4 \\ x &= 2 \end{align}$$ If we use this value of x in either equation to find y, we get y = 0. The point of intersection is |

12. |
$$ \begin{align} 13x - 11y &= 1 \\ x - 3y &= 2 \end{align}$$ ## SolutionThis looks like a good one for substitution. Solving the second equation for x gives $$ \begin{align} 13(3y + 2) - 11y &= 1 \\ 39y + 26 - 11y &= 1 \\ 28y &= -25 \\ y &= \frac{-25}{28} \end{align}$$ Now substitute y into the simpler of our two equations to get x: $$ \begin{align} x &= 3y + 2 \\ x &= 3 \left( \frac{-25}{28} \right) + \frac{56}{28} \\ x &= \frac{-75 + 56}{28} \\ x &= \frac{-19}{28} \end{align}$$ The point of intersection is $(\frac{-19}{28}, \frac{-25}{28}).$ |

There are many pretty cool applications of solving linear systems. We'll do some examples here and finish with some practice problems.

A school organization collects quarters and dimes to raise funds for a charity. In all, \$147.20 is collected. If a total of 740 coins were collected, how many quarters and dimes were collected?

**Solution**

First, the total number of coins is 740, that means

$$q + d = 740$$

Second, the values of the coins (in dollars), must add to \$147.2. We write that like this:

$$0.25q + 0.10d = 147.2$$

Now we can solve these two equations using one of the techniques we've learned. Let's use substitution, solving the first equation for d and substituting that for d in the second:

$$ \begin{align} d &= 740 - q \\[2pt] 0.25 q + 0.10(740 - q) &= 147.2 \\[2pt] 0.25 q + 74 - 0.1q &= 147.2 \\[2pt] 0.15 q &= 73.2 \\[2pt] q &= 488 \; \text{ quarters} \end{align}$$

Now it's easy to find the number of dimes from the first equation:

$$d = 740 - 488 = 252 \; \text{ dimes}$$

So the money taken in consisted of **488 quarters and 252 dimes**, for a total of \$147.2.

A sum of \$5,400 is invested in two accounts. Part of it was invested in an account bearing 4.25% interest annually, and the other in a bond which yielded 3.0% at the end of the year. If the interest earned at the end of the year was $204.20, how much money was initially invested in each account?

**Solution**

First, we know that the total amount is

$$a + b = 5400.0$$

Now the interest earned on the bank investment would be $0.0425 a,$ and the interest earned on the bond would be $0.03 b.$ So we can write a linear equation for the interest earned:

$$0.0425 a + 0.03 b = 204.20$$

Now we can solve these two equations. Substitution looks like the way to go here, too. Let's solve the first equation for b, then plug that expression into the second.

$$ \begin{align} b &= 5400 - a \\[2pt] 0.0425 a + 0.03(5400 - a) &= 240.2 \\[2pt] 0.0425 a + 162 - 0.03 a &= 240.2 \\[2pt] 0.125 a &= 78.2 \\[2pt] a &= \$ 625.60 \end{align}$$

Now it's easy to find the amount invested in bonds from the first equation

$$b = 5400 - 625.60 = \$ 4,774.40$$

So **\$625.60 was invested in the bank account**, and the remaining **\$4774.40 was invested in the bond**.

A 1% saline (salt-water) solution and a 5% saline solution are to be mixed to yield 100 mL of a 2.4% solution. How many milliliters (mL) of each should be mixed to achieve the desired solution?

In this problem, we have two variables. Let's let x be the amount (in mL) of 1% saline and y be the amount of 5% saline. We also have two pieces of information. The first is that the total number of mL has to be 100. The second is that the total concentration (salinity) has to be 2.4%.

The first part is easy, $x + y = 100.$

The second part is a little more tricky because involves percents. The final concentration is the amount of salt (which we find by multiplying the percent saline by the volume – x or y), divided by the total volume, 100 mL.

$$\frac{0.01x + 0.05y}{100} = 0.024$$

We can multiply both sides of that equation by 100 to get

$$0.01x + 0.05y = 2.4$$

So our system of two equations is

$$ \begin{align} x + y &= 100\\[2pt] 0.01x + 0.05y &= 2.4 \end{align}$$

Substitution looks like our best bet here: If $x + y = 100,$ then $x = 100 - y.$ So our substitution is:

$$ \begin{align} 0.01(100 - y) + 0.05y &= 2.4 \\[2pt] 1 - 0.01y + 0.05y &= 2.4 \\[2pt] 1 + 0.04y &= 2.4 \\[2pt] 0.04y &= 1.4 \\[2pt] y &= \frac{1.4}{0.04} = 35 \; mL \end{align}$$

Now $x = 100 - y = 100 - 35 = 65 \; mL$

So the recipe for our new solution is **65 mL of the 1% solution** and **35 mL of the 5% solution**.

13. |
A pile of cash contains \$725.00 and consists of 72 bills in denominations of \$20 and \$5 only. How many of each \$20 and \$5 bill are in the pile. Also, can I keep that cash? ## SolutionLet's let the number of fives be We can solve this by substitution: $$ \begin{align} f + t &= 73 \; \rightarrow \; f = 73 - t \\ 5(73 - t) + 20t &= 725 \\ 365 - 5t + 20t &= 725 \\ 15t &= 365 \\ t &= \frac{365}{15} = 24 \; \text{ twenties} \end{align}$$ Then the number of fives is $$73 - 24 = 49 \; \text{ fives}$$ So the pile contains 49 fives and 24 twenties for a total of 73 bills summing to \$ 725.00 |

14. |
A rectangular sports field has a perimeter of 320 m. Its length is 20 m longer than its width. Calculate the dimensions of the field. ## SolutionThis one isn't exactly linear, but we can still do it. If you don't know what completing the square is, just skip this problem. Our equations are (from the perimeter and area, respectively), $$ \begin{align} 2x + 2y &= 320 \; \longrightarrow \; x + y = 60 \\[2pt] xy &= 5000 \end{align}$$ Substituting the former into the latter gives: $$ \begin{align} x(160 - x) &= 5000 \\[2pt] -x^2 + 160 x &= 5000 \\[2pt] x^2 - 160 x &= -5000 \\[2pt] x^2 - 160 x + 80^2 &= 6400 - 5000 \leftarrow \text{compl. the square} \\[2pt] (x - 80)^2 &= 1400 \\[2pt] x &= 80 ± \sqrt{1400} \\[2pt] x &= 80 ± 37.42 \\[2pt] x &= 117.42, \; 42.58 \end{align}$$ These happen to add to 160, so they are our x and y. They yield a perimeter of 320 m and an area of 5000 m The dimensions of the field are 117.42 m × 42.58 m. |

15. |
A coffee blender has some expensive coffee (\$9 per pound) that doesn't sell well. She wants to blend it with some \$4 per pound coffee to make a mixture that can sell for \$5.50 per pound. How many pound of each coffee will she need in order to make 50 pounds of the \$5.50/lb. coffee? ## SolutionLet $$ \begin{align} e + c &= 50 \\[2pt] 9e + 4c &= 5.5(50) \; \leftarrow \; c = 50 - e \\[2pt] 9e + 4(50 - e) &= 275 \\[2pt] 9e + 200 - 4e &= 275 \\[2pt] 5e &= 75 \\[2pt] e &= 15 \; lbs. \\[2pt] c &= 35 \; lbs. \end{align}$$ The blender would mix 15 lb. of the expensive coffee with 35 lb. of the cheap to make 50 lb. of a blend that could sell for \$5.50 per pound. |

16. |
Let's say we need 250 mL of a 4% glucose solution, but all we have available are stock solutions of 2% glucose and 25% glucose. How many mL of each should we mix to get 250 mL of the 4% solution? ## SolutionLet's make w the number of milliliters (mL) of our weak (2%) solution and s the number of mL of the strong (25%) solution. The percent concentration (as a decimal) is the number of mL of glucose divided by the number of mL of the solution (250 mL). We have two constraints or pieces of information. They are: $$ \begin{align} w + s &= 250 \\[2pt] \frac{0.25 s + 0.02 w}{250} &= 0.04 \; \text{ or} \\[2pt] 0.25 s + 0.02 w &= 10 \end{align}$$ Now if we substitute $$ \begin{align} 0.25 s + 0.02(250 - s) &= 10 \\[2pt] 0.25 s + 5 - 0.02 s &= 10 \\[2pt] 0.23 s &= 5 \\[2pt] s &= 21.74 \; mL \end{align}$$ then w = 250 - 21.74 = 228.26 mL The solution is made by adding 21.74 mL of the strong (25%) solution to 228.26 mL of the weak (2%) solution. |

17. |
In 2019, two groups of people bought tickets for Boston Redsox and Seattle Mariners games. Group-1 bought 4 'Sox tickets and 2 Mariners tickets for a total of \$500.00. Group-2 bought 3 'Sox tickets and 1 Mariners ticket for a total of \$344.00. Based on this information, What were the average costs (assuming these are average ticket prices) of Redsox and Mariners tickets in 2019? ## SolutionIn this question, we have two equations of the same type, both sums of costs of tickets of two types. Let $$ \begin{align} 4s + 2m &= 500.00 \\ 3s + m &= 344.00 \end{align}$$ This system is best solved by elimination. We'll multiply the bottom equation by -2 and add: $$ \begin{matrix} 4s + 2m = 500.00\phantom{0} && \\ \underline{3s + m = 344.00} && \leftarrow \times -2 \\ 4s + 2m = 500.00 && \\ \underline{-2(3s + m = 344)\phantom{000}} && \\ 4s + 2m = 500\phantom{00} && \\ \underline{-6s - 2m = -688\phantom{00}} && \leftarrow \text{ add} \\ \phantom{00}-2s = -188 && \\ \phantom{000000}s = \$ 94.00 && \end{matrix}$$ Then using either of the original equations, we get $m = \$62.00.$ Redsox tickets: \$94.00, Mariners tickets: \$62.00 |

18. |
An investor invests \$10,000 in two investments, a real-estate trust (REIT) paying 5.5% interest annually, and a mutual fund paying 7.5% annually. At the end of the year, the combined new balance of the two accounts is \$10,532.00. How much money did the the investor begin the year with in each account? ## SolutionLet $$ \begin{align} r + m &= 1000.00 \\[2pt] 0.055r + 0.075m &= 532 \end{align}$$ where 532.00 is the amount of interest earned by the end of the year (the amount over the initial \$10,000 investment. Now we can solve the system by substitution: $$ \begin{align} r + m &= 1000.00 \rightarrow r = 1000 - m \\[2pt] 0.055(1000 - m) + 0.075m &= 532 \leftarrow \text{ substitute} \\[2pt] 55 - 0.055m + 0.075m &= 532 \\[2pt] 0.02m &= 477 \\[2pt] m &= \frac{477}{0.2} = \$2385.00 \\[2pt] r = 1000 - 2385 &= \$7615.00 \end{align}$$ The original investment was \$2,385.00 in the mutual find and \$7,615.00 in the REIT. |

19. |
A car travels 300 miles on a highway in the same time that a loaded truck travels 280 miles. If the speed of the car is, on average, 5 mi./h faster than the truck's, calculate the speeds of the car and truck. ## SolutionLet $$s = \frac{d}{t} \longrightarrow t = \frac{d}{s},$$ where s = speed, d = distance and t = time. Because the times are identical, we can set d/s equal for both vehicles: $$ \begin{align} \frac{300}{C} = \frac{280}{T} \\[4pt] \frac{300}{T + 5} = \frac{280}{T} \\[4pt] 300T = 280T + 1400 \\[2pt] 20T = 1400 \\[4pt] T = \frac{1400}{20} = 70 \; \text{mi./h} \\[2pt] C = 75 \; \text{mi./h} \end{align}$$ The speed of the car was 75 mi/h and the speed of the truck was 70 mi/h. |

20. |
Astronomy club members want to order T-shirts with the club logo on them. One company charges \$8.50 per shirt with an initial set-up fee of \$40.00. The other company charges \$9.25 per shirt with an initial fee of \$24.00. For what number of T-shirts would the total cost be the same? ## SolutionIf we let $$ \begin{align} C_1 &= 8.5n + 40 \\[2pt] C_2 &= 9.25n + 24 \end{align}$$ Now we're interested in what n is when C $$ \begin{align} 8.5n + 40 &= 9.25n + 24 \\[2pt] 16 &= 0.75 \, n \\[2pt] n &= \frac{16}{0.75} = 21.33 \; \text{ shirts} \end{align}$$ So as long as 22 or more shirts are ordered, both companies are the same (in terms of cost, anyway). If fewer than 22 are ordered, company 2 is cheaper. |

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