Lesson 14Solving More Systems

Let’s solve systems of equations.

Learning Targets:

  • I can use the structure of equations to help me figure out how many solutions a system of equations has.

14.1 Algebra Talk: Solving Systems Mentally

Solve these without writing anything down:

\begin{cases} x=5\\ y=x-7 \end{cases}

\begin{cases} y=4\\ y=x+3 \end{cases}

\begin{cases} x=8\\ y=\text-11 \end{cases}

14.2 Challenge Yourself

Here are a lot of systems of equations:

\begin{cases} y= 4 \\ x=\text-5y+6 \end{cases}

\begin{cases} y= 7 \\ x=3y-4 \end{cases}

\begin{cases} y= \frac{3}{2}x+7 \\ x=\text-4 \end{cases}

\begin{cases} y= \text-3x+10 \\ y=\text-2x+6 \end{cases}

\begin{cases} y= \text-3x-5 \\ y=4x+30 \end{cases}

\begin{cases} y= 3x-2 \\ y=\text-2x+8 \end{cases}

\begin{cases} y= 3x \\ x=\text-2y+56 \end{cases}

\begin{cases} x=2y-15 \\ y= \text-2x \end{cases}

\begin{cases} 3x+4y=10 \\ x=2y \end{cases}

\begin{cases} y= 3x+2 \\ 2x+y = 47 \end{cases}

\begin{cases} y= \text-2x+5 \\ 2x+3y = 31 \end{cases}

\begin{cases} x+y=10 \\ x=2y +1 \end{cases}

  1. Without solving, identify 3 systems that you think would be the least difficult to solve and 3 systems that you think would be the most difficult to solve. Be prepared to explain your reasoning.
  2. Choose 4 systems to solve. At least one should be from your "least difficult" list and one should be from your "most difficult" list.

14.3 Five Does Not Equal Seven

Tyler was looking at this system of equations:

\begin{cases} x + y = 5\\x + y = 7 \end{cases}

He said,

“Just looking at the system, I can see it has no solution. If you add two numbers, that sum can’t be equal to two different numbers.”

Do you agree with Tyler?

Are you ready for more?

In rectangle ABCD , side AB is 8 centimeters and side BC is 6 centimeters. F is a point on BC and E is a point on AB . The area of triangle DFC is 20 square centimeters, and the area of triangle DEF is 16 square centimeters. What is the area of triangle AED

Lesson 14 Summary

When we have a system of linear equations where one of the equations is of the form y = \text{[stuff]} or x=\text{[stuff]} , we can solve it algebraically by using a technique called substitution. The basic idea is to replace a variable with an expression it is equal to (so the expression is like a substitute for the variable). For example, let's start with the system:

\begin{cases} y = 5x\\2x - y = 9 \end{cases}

Since we know that y = 5x , we can substitute 5x for y in the equation 2x - y = 9 ,

2x - (5x) = 9,

and then solve the equation for x ,

x =\text -3.

We can find y using either equation. Using the first one: y = 5 \boldcdot \text-3 . So

(\text-3,\text -15)

is the solution to this system. We can verify this by looking at the graphs of the equations in the system:

Sure enough! They intersect at (\text-3, \text-15) .

We didn't know it at the time, but we were actually using substitution in the last lesson as well. In that lesson, we looked at the system

\begin{cases} y = 2x + 6 \\ y  = \text-3x - 4 \end{cases}

and we substituted 2x+6 for y into the second equation to get 2x+6=\text-3x-4 . Go back and check for yourself!

Lesson 14 Practice Problems

  1. Solve: \begin{cases} y=6x \\ 4x+y=7 \\ \end{cases}

  2. Solve: \begin{cases} y=3x \\ x=\text-2y+70 \\ \end{cases}

  3. Which equation, together with y=\text-1.5x+3 , makes a system with one solution?

    1. y=\text-1.5x+6

    2. y=\text-1.5x

    3. 2y=\text-3x+6

    4. 2y+3x=6

    5. y=\text-2x+3

  4. The system x-6y=4 , 3x-18y=4 has no solution.

    1. Change one constant or coefficient to make a new system with one solution.

    2. Change one constant or coefficient to make a new system with an infinite number of solutions.

  5. Match each graph to its equation.

    1. y=2x+3
    2. y=\text-2x+3
    3. y=2x-3
    4. y=\text-2x-3
  6. Here are two points: (\text-3,4) , (1,7) . What is the slope of the line between them?

    1. \frac43
    2. \frac34
    3. \frac16
    4. \frac23