Have you ever been to the Omni theater? Take a look at this dilemma.

Kelly loved the Omni presentation on the rainforest so much that she decided to go and see it again. She asked Tyler if he wanted to go with her on Saturday afternoon.

“Do you want to go with me?” Kelly asked.

“Sure, but I have karate first, so I will have to meet you there. What time is the show?” Tyler asked.

“The show starts at 2:30 pm. I’m going to leave at one o’clock so I can look around,” Kelly said.

“Well, I don’t get done karate until then, so I probably won’t leave until 2:00 pm,” Tyler said.

On Saturday, the two went about their day and both left for the museum. Kelly’s Mom tends to drive slowly and cautiously, so she was traveling on city streets at 45 mph. Tyler’s karate class is downtown, so he could take the highway to get to the Omni theater and his Dad drove at an average of 55 mph. Will the two catch up to each other?

**This is a problem about equations and systems of equations. You will need to solve a system of equations to figure this one out. To figure this out, you would need to find a solution that will work for both equations. You will learn all about it in this Concept.**

### Guidance

Let’s recall that in a system of equations, we are looking for the same \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values that make all equations true. So, the solution to the system

\begin{align*}3y &=x-2 \\ y-x &=4 \end{align*}

is the ordered pair \begin{align*}(x, y)\end{align*} that makes the first and the second equation true. In other words, \begin{align*}x\end{align*} in the first equation equals \begin{align*}x\end{align*} in the second equation and \begin{align*}y\end{align*} in the first equation equals \begin{align*}y\end{align*} in the second equation. Now, look at the second equation, \begin{align*}y-x=4\end{align*}. It is simple to solve for the \begin{align*}y\end{align*} value—add \begin{align*}x\end{align*} to both sides. So \begin{align*}y=x+4\end{align*}. Well, if \begin{align*}y\end{align*} is the same in both equations and \begin{align*}y=x+4\end{align*}, then we can substitute \begin{align*}x + 4\end{align*} in the place of \begin{align*}y\end{align*} in the first equation. That is why this is called the *substitution method*.

Now that we have substituted, we can solve this equation because it has a single variable.

\begin{align*}3(x+4) &= x-2 \\ 3x+12 &=x-2 \\ 2x+12 &=-2 \\ 2x &=-14 \\ x &=-7 \end{align*}

If \begin{align*}x = -7\end{align*}, substitute again to find \begin{align*}y\end{align*}:

\begin{align*} y &=x+4 \\ y &=-7+4 \\ y &=-3 \end{align*}

**Our solution, then, is (-7, -3).**

Solve each system by using substitution.

#### Example A

\begin{align*}2y &=x+4 \\ y &= 3x\end{align*}

**Solution: \begin{align*}x = \frac{4}{5}, y= 2 \frac{2}{5}\end{align*}**

#### Example B

\begin{align*}3y &=x-22 \\ y &=4x \end{align*}

**Solution: \begin{align*}x = -2,y = -8\end{align*}**

#### Example C

\begin{align*}6y &=x-34 \\ y &=3x \end{align*}

**Solution: \begin{align*}x = 2, y = 6\end{align*}**

Now let's go back to the dilemma from the beginning of the Concept.

**The first equation we can write is to represent Tyler’s time. His Dad is traveling 55 mph. Therefore his distance is a function of speed and time.**

\begin{align*}d = 55t\end{align*}

**Kelly left one hour before Tyler did. She is traveling 45 miles per hour. Therefore, the speed times Tyler’s time plus one hour equals Kelly’s time.**

\begin{align*}d = 45(t + 1)\end{align*}

**Now see if there is a solution that will work for both equations. We can try to solve this by using substitution.**

\begin{align*}55t &= 45(t + 1) \\ 55t &= 45t + 1 \\ 55t - 45t &= 1 \\ 10t &= 1 \\ t &= \frac{1}{10} \end{align*}

**Now we go to Tyler.**

\begin{align*}d &= 55\left ( \frac{1}{10} \right ) \\ d &= 5.5 \end{align*}

**The solution could be the following values for \begin{align*}d\end{align*}** *and***\begin{align*}t\end{align*}. However, when you substitute those values into both equations, the solution does not work. Therefore there isn’t a solution for this system, and the Kelly and Tyler won’t meet up while driving.**

### Vocabulary

- System of Equations
- Two or more equations at the same time. The solution will be the ordered pair that works for both equations.

### Guided Practice

Here is one for you to try on your own.

Anglica’s mother leaves to visit her grandmother for her birthday. Her grandmother lives 450 miles away and her mother drives at an average of 60mph. Three hours later, Angelica’s step-father notices that her mother forgot the gift for her grandmother so he decides to try to catch up to her. If he drives at an average of 50mph, will he catch up to her before she gets to her grandmother’s house?

Write a system of equations to model the situation and solve using the substitution method.

**Solution**

First, we can figure out how long it will take Anglica's mother to reacher her grandmother's house given the distance.

\begin{align*}d = rt\end{align*}

She is driving 50 mph, and her grandmother lives 450 miles away.

\begin{align*}480 &= 60t \\ 8 &= t\end{align*}

**It will take her eight hours.**

Now let's write and solve a system of equations to see if her step-father will catch up before her mother arrives.

\begin{align*}d = 60t\end{align*} This is the mother's driving.

\begin{align*}d = 50t + 3\end{align*} This is the step-father's driving.

\begin{align*}60t=50t+3\end{align*}

Now we solve the equation for t, the time.

\begin{align*}60t-50t &= 3 \\ 10t &= 3 \\ t &= \frac{3}{10}\end{align*}

Now substitute this back into the mother's time.

\begin{align*}60 \times \frac{3}{10}\end{align*}

\begin{align*}18\end{align*}

**It will take the step-father 18 hours to catch up. He will not arrive before Anglica's mother.**

### Video Review

Solving Linear Systems by Substitution

### Practice

Directions: Solve each linear system by using substitution.

- .

- \begin{align*} y &= 8 \\ 2y + 2x &= 2 \end{align*}

- .

- \begin{align*} x &= 6 \\ 2y + 2x &= 2 \end{align*}

- .

- \begin{align*} 2x & = 8 \\ 2y + x &= 10 \end{align*}

- .

- \begin{align*} 3y &= 9 \\ y + 2x &= 11 \end{align*}

- .

- \begin{align*} y -8 &= 8 \\ 2y + x &= 20 \end{align*}

- .

- \begin{align*} 4y &= 8 \\ y - 2x &= 8 \end{align*}

- .

- \begin{align*} x -2 &= 4 \\ 4x + y &= 12 \end{align*}

- .

- \begin{align*} y &= x+8 \\ y + 2x &= 11 \end{align*}

- .

- \begin{align*} 2y +8 &= 12 \\ y + 2x &= 20 \end{align*}

- .

- \begin{align*} y -3 &= 6 \\ 3y + 3x &= 9 \end{align*}

- .

- \begin{align*} 4y -1 &= 11 \\ y - 4x &= 5 \end{align*}

- .

- \begin{align*} 2y -8 &= 8 \\ 2y + 2x &= 2 \end{align*}

- .

- \begin{align*} 4x + y &= -2 \\ -2x - 3y &= 1 \end{align*}

- .

- \begin{align*} y &= 2x \\ 6x - y &= 8 \end{align*}

- .

- \begin{align*} x + 4y &= -6 \\ 2x + 10y &= -6 \end{align*}