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As a reference see: "What Do Linear Quadratic Systems Look Like?"
A linear quadratic system is a system containing one linear equation and one quadratic equation
which, for Algebra 1, will be one straight line and one parabola.
 Algebraic Solutions |
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straight line: y = mx + b
parabola: y = ax2 + bx + c; a ≠ 0
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When working with linear-quadratic systems, we will be solving the linear equation for one of its variables (preferably "y"), and substituting that value into the quadratic equation.
Solve this linear-quadratic system of equations algebraically and check your solution:
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y = x2- 6x + 3 (parabola)
y = -2x + 3 (straight line)
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1. Solve for one of the variables in the linear equation.
Note: In this example, this process is already done for us, since y = -2x + 3. |
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2. Substitute this value into the quadratic equation, and solve the resulting equation.
• Substitute -2x + 3 for y in the quadratic equation.
• Subtract 3 from both sides; then add 2x to both sides.
• Factor.
• Set each factor equal to zero and solve.
You now have TWO values for x. This tells you that there may be two possible solutions.
TWO SOLUTIONS.
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y = x2 - 6x + 3
-2x + 3 = x2 - 6x + 3
0 = x2 - 4x + 0
0 = x(x - 4)
x = 0; x - 4 = 0
x = 0; x = 4 |
3. Find the corresponding values for y. Substitute each value into the linear equation in place of x.
Yes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
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x = 0
y = -2(0) + 3
y = 3
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Solution 1:
x = 0; y = 3
or (0, 3) |
x = 4
y = -2(4) + 3
y = -5 |
Solution 2:
x = 4; y = -5
or (4, -5) |
4. Check: Be sure to check BOTH solutions in both equations.
Solution 1:
x = 0; y = 3
y = -2x + 3
3 = -2(0) + 3
3 = 3 check!
y = x2 - 6x + 3
3 = (0)2 - 6(0) + 3
3 = 3 check! |
Solution 2:
x = 4; y = -5
y = -2x + 3
-5 = -2(4) + 3
-5 = -5 check!
y = x2 - 6x + 3
-5 = (4)2 - 6(4) + 3
-5 = 16 - 24 + 3
-5 = -5 check! |
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| 5. State the final solutions. The solutions may be stated as the set {(0, 3), (4, -5)} |
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Notice that in this example both equations started out set equal to "y".
The "y =" value from one equation was then substitued into the other equation to allow for a solution.
You can think of this process of starting out with both equations set equal to "y"
as a fast and easy "solution method".
Solve this linear-quadratic system of equations algebraically and check your solution: |
y = x2- 6x + 3 (parabola)
2x - y = 13 (straight line) |
1. Solve for one of the variables in the linear equation, preferrably "y".
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2. Substitute this value into the quadratic equation, and solve the resulting equation.
• Substitute 2x - 13 for y in the quadratic equation.
• Add 13 to both sides; then subtract 2x from both sides.
• Factor.
• Set each factor equal to zero and solve.
You now have ONE value for x. This tells you that there may be only one solution.
ONE SOLUTION
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y = x2 - 6x + 3
2x - 13 = x2 - 6x + 3
0 = x2 - 8x + 16
0 = (x - 4)(x - 4)
x - 4 = 0; x - 4 = 0
x = 4; x = 4 |
3. Find the corresponding value for y. Substitute the value into the linear equation in place of x.
Yes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
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x = 4
y = 2(4) - 13
y = -5 |
Solution 1:
x = 4; y = -5
or (4, -5) |
4. Check: Be sure to check the solution in both equations.
Solution 1:
x = 4; y = -5
y = 2x - 13
-5 = 2(4) - 13
-5 = -5 check!
y = x2 - 6x + 3
-5 = (4)2 - 6(4) + 3
-5 = 16 - 24 + 3
-5 = -5 check! |
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5. State the final solution. The solution may be stated as (4, -5) or {(4, -5)} |
FYI: This next example involves a stratight line with a circle.
This generally is not a topic for Algebra 1,
but may be seen in Algebra I Honors Level courses.
Solve this linear-quadratic system of equations algebraically and check your solution:
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x2+ y2 = 9 (circle)
x - y = 3 (straight line)
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1. Solve for one of the variables in the linear equation.
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2. Substitute this value into the quadratic equation, and solve the resulting equation.
• Substitute x - 3 for y in the quadratic equation.
• Expand (x - 3)2
• Combine terms.
• Factor.
• Set each factor equal to zero and solve.
You now have TWO values for x. This tells you that there may be two possible solutions.
TWO SOLUTIONS.
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x2 + (x - 3)2 = 9
x2 + x2 - 6x + 9 = 9
2x2 - 6x = 0
2x(x - 3) = 0
2x = 0; x - 3 = 0
x = 0; x = 3 |
3. Find the corresponding values for y. Substitute each value into the linear equation in place of x.
Yes, you could substitute in the quadratic equation, but substituting into the linear equation will be easier.
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x = 0
y = 0 - 3
y = -3
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Solution 1:
x = 0; y = -3
or (0, -3) |
x = 3
y = 3 - 3
y = 0 |
Solution 2:
x = 3; y = 0
or (3, 0) |
4. Check: Be sure to check BOTH solutions in both equations.
Solution 1:
x = 0; y = -3
y = x - 3
-3 = 0 - 3
-3 = -3 check!
x2 + y2 = 9
02 + (-3)2 = 9
9 = 9 check! |
Solution 2:
x = 3; y = 0
y = x - 3
0 = 3 - 3
0 = 0 check!
x2 + y2 = 9
32 + 02 = 9
9 + 0 = 9
9 = 9 check! |
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| 5. State the final solutions. The solutions may be stated as the set {(0, -3), (3, 0)} |
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Yes, this is an algebraic solution method. But, you can still use your graphing calculator to double CHECK your solution. Click the calculator at the right to see how to use the calculator with linear quadratic systems.
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For calculator help with linear quadratic systems
click here. |
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