Solving Systems of Equations
Linear and Quadratic
Activator: Write down what you remember about quadratics.
Write down what you remember about linear functions.
Activity 1:
Graph the quadratic: y = x2 – 4x + 3
1. Write down some characteristics of
the graph.
Graph the linear equation: y = 3
2. In how many places do the graphs intersect? Do you think a quadratic equation and a linear equation
will always intersect this many times? Explain.
3. Write the coordinates of the intersections. ________________________________
4. Let’s see how good your guesses are by solving this algebraically! Do you remember the substitution
method from when you were solving systems of linear equations? Write down what you remember.
5. Let’s apply it to our problem. We can substitute the linear equation into the quadratic equation for y.
Why can we do this?
6. Once you have done that, you have a quadratic equation in only one variable, x, that you can solve for!
Because you are solving a quadratic, how many solutions can you get?
7. How do you solve a quadratic equation?
8. Write the new quadratic equation here. _____________________________________________
9. Show how to solve the quadratic.
10. You have solved the quadratic for x. Is that your final answer? No! You also have to find the
corresponding y-values!! How can we find the corresponding y-values?
11. Show how to find the y-values.
12. What are the solutions to this problem? ______________________________
13. Do your solutions match your answers from #3? Are you exact or close?
14. If they are exact, why do you think you guessed so well? If you are close, why do you think you only
got close? If you are not even close, why do you think that happened?
Activity 2:
Graph the quadratic: y = x2 – 4x + 3
Graph the linear equation: y = x – 1
1. Write the coordinates of the intersections.
2. After combining the two equations, write
the new quadratic equation in only one
variable.
3. Solve the quadratic for the x-values.
5. Write the solutions as ordered pairs. ___________________________
Activity 3:
Find the solutions to this system:
y = x2 – 4x + 3
𝑦=
−1
2
𝑥+
3
2
4. Find the corresponding y-values.
Extension:
Find the solutions to this system:
(x – 3)2 + (y – 3)2 = 25
𝑦 = 𝑥 +5
Summarizer: Ticket out the Door
Today we solved a system of equations where one equation was linear and the other quadratic. We got
two solutions for all of our answers. Explain the cases where we might only have one solution AND no
solutions. Draw a picture of each case.
Solving Systems of Equations – Teacher Notes
Linear and Quadratic
Activator: Write down what you remember about quadratics. x2 term; graph a parabola; can have at
most 2 zeroes, solutions, or x-intercepts; have 1 y-intercept; open up or down depending on leading
coefficient; have a max or min depending on if it opens up or down;…
Write down what you remember about linear functions. Can be written in slope-intercept form; y = mx + b;
have + / – slope; “intercept” is y-intercept; horizontal lines are linear functions; vertical lines are not linear
functions; horizontal lines can be written as y = c; vertical lines can be written as x = d; …
**You should point out that when you were covering the unit on quadratics that you solved quadratic
equations by setting them equal to zero and solving for x. This tells you the “solutions” to the quadratic that
we also call “zeroes” or “roots.” When going through that process, you are actually finding the points of
intersection of the quadratic (the parabola) and the specific line that makes up the x-axis, which is y = 0.
Today, we are going to find those points of intersection when the line is NOT y = 0.
Activity 1:
Graph the quadratic: y = x2 – 4x + 3
1. Write down some characteristics of
the graph.
Has roots 1 & 3 which means it factors into
y = (x – 1)(x – 3); opens up; has min at (2, -1);
has y-int at (0, 3); …
Graph the linear equation: y = 3
2. In how many places do the graphs intersect? Do you think a quadratic equation and a linear equation
will always intersect this many times? Explain.
They intersect in two places, but they could intersect in only one place OR in zero places.
3. Write the coordinates of the intersections. (0, 3) & (4, 3)
4. Let’s see how good your guesses are by solving this algebraically! Do you remember the substitution
method from when you were solving systems of linear equations? Write down what you remember.
You can solve one of the equations for one of the variables and substitute this into the second equation for
the same variable. For instance, if you solve one equation for y, you can substitute what it equals into the
other equation for y.
5. Let’s apply it to our problem. We can substitute the linear equation into the quadratic equation for y.
Why can we do this?
Because both of our equations are y = something so we can just set them equal to each other as we substitute
one equation in for the other.
6. Once you have done that, you have a quadratic equation in only one variable, x, that you can solve for!
Because you are solving a quadratic, how many solutions can you get?
We can get two, one, or no solutions.
7. How do you solve a quadratic equation?
You can factor it & set each factor equal to zero; you can complete the square; you can graph it and look
where the graph crosses the x-axis; or you can use the quadratic formula.
8. Write the new quadratic equation here. 3 = x2 – 4x + 3
9. Show how to solve the quadratic.
3 = x2 – 4x + 3
0 = x2 – 4x
0 = x (x – 4)
0 = x AND 0 = x – 4  x = 4
10. You have solved the quadratic for x. Is that your final answer? No! You also have to find the
corresponding y-values!! How can we find the corresponding y-values?
Substitute the x-values that we found into either equation and solve for y.
11. Show how to find the y-values.
For x = 0, y = 3 AND for x = 4, y = 3.
12. What are the solutions to this problem as ordered pairs? (0, 3) & (4, 3)
13. Do your solutions match your answers from #3? Are you exact or close?
YES! Mine are exact!!
14. If they are exact, why do you think you guessed so well? If you are close, why do you think you only
got close? If you are not even close, why do you think that happened?
My intersections on my graph looked like they were whole numbers, which made guessing very easy. If the
answers had been fractions or irrational numbers, I would have only been able to get an approximation or
close!
Activity 2:
Graph the quadratic: y = x2 – 4x + 3
Graph the linear equation: y = x – 1
4. Write the coordinates of the intersections.
(1, 0) & (4, 3)
5. After combining the two equations, write
the new quadratic equation in only one
variable.
x – 1 = x2 – 4x + 3
6. Solve the quadratic for the x-values.
x – 1 = x2 – 4x + 3
0 = x2 – 5x + 4
0 = (x – 1)(x – 4)
x – 1 = 0 and x – 4 = 0
x=1
and
x=4
5. Write the solutions as ordered pairs. (1, 0) & (4, 3)
Activity 3:
Find the solutions to this system:
y = x2 – 4x + 3
𝑦=
−1
2
𝑥+
3
2
(3, 0) & (0.5, 1.25)
4. Find the corresponding y-values.
Using y = x – 1, when x = 1,
y = 1 – 1 = 0.
When x = 4, y = 4 – 1 = 3.
Extension:
Find the solutions to this system:
(x – 3)2 + (y – 3)2 = 25
𝑦 = 𝑥 +5
(-2, 3) & (3, 8)
Summarizer: Ticket out the Door
Today we solved a system of equations where one equation was linear and the other quadratic. We got
two solutions for all of our answers. Explain the cases where we might only have one solution AND no
solutions. Draw a picture of each case.
If there is only one solution, the line would be tangent to the parabola. If there is no solution, then the line
doesn’t intersect the parabola at all.