Solving Linear Quadratic Systems Algebraically
 Topic Index | Algebra Index | Regents Exam Prep Center

Let's look at how to solve a linear quadratic system of equations algebraically.

Example 1:

Solve  this system of equations algebraically:
                    y = x2 - x - 6    (quadratic equation of form y = ax2 + bx + c:  parabola)
                    y = 2x - 2         (linear equation of form y = mx + b)

First, we solve for one 
of the variables in the
   linear equation.		 y = 2x - 2

 

 Since this is already done for us in this example, we can go to the next step.
Next, we substitute for 
that variable in the 
quadratic equation, and 
solve the resulting 
equation.		 y = x2 - x - 6
 2x - 2 = x2 - x - 6
 2x = x2 - x - 4

0 = x2 - 3x - 4

0 =(x - 4)(x + 1)

x - 4 = 0      x + 1 =0
x = 4           x = -1

  

Add 2 to both sides.
Subtract 2x from both sides.

Factor.
 


Set each factor = 0 and solve.
We now have two values for x, but we still need to find the corresponding values for y.
We find the y-values by  
substituting each value 
of x into the linear  
equation.
y = 2x - 2

Check 4
y = 2(4) - 2
y = 8 - 2
y = 6		 (4, 6)
y = 2x - 2

Check -1
y = 2(-1) - 2
y = -2 - 2
y = -4		 (-1, -4)
Now we have 2 possible solutions for the system:  (4,6) and (-1,-4).  We need to check each solution in each equation.
          
Check#1:  (4, 6)
y = x2 - x - 6

6 = (4)2 - 4 - 6
6 = 16 - 4 - 6
6 = 6   it checks !

y = 2x - 2

6 = 2(4) - 2
6 = 8 - 2
6 = 6   it also checks !
Check#2:   (-1, -4)
y = x2 - x - 6

-4 = (-1)2 - (-1) - 6
-4 = 1 + 1 - 6
-4 = -4   it checks !

y = 2x - 2

-4 = 2(-1) - 2
-4 = -2 - 2
-4 = -4   it also checks !
We finally have our solution set for this linear quadratic system.

{(4, 6), (-1, -4)}
 

 

Example 2:

Solve  this system of equations algebraically:
                   x2 + y2 = 26   (quadratic equation of form x2 + y2 = r2:  circle)
                    x - y = 6        (linear equation)

First we solve for x in the linear equation. x - y = 6
x = y + 6

 
Add y to both sides
Now substitute this value
of x into the quadratic
 equation replacing the x.
 Solve the resulting equation.		 x2 + y2 = 26
(y + 6)2 + y2 = 26
y2 + 12y + 36 + y2 = 26
2y2 + 12y + 36 = 26
 2y2 + 12y + 10 = 0

y2 + 6y + 5 = 0

(y + 5)(y +1)=0
y + 5=0    y + 1=0
y = -5       y = -1

 

Expand (y + 6)2
Combine similar terms.

 Divide each term by 2.
Factor.
 Set each factor = 0.


 
Next, find the values of y by substituting in the linear equation.

x - y = 6

x - (-5) = 6
x + 5 = 6
x = 1*
 (1,-5)

x - y = 6

x - (-1) = 6
x + 1 = 6
x = 5*
 (5,-1)

* Remember to write the x-values first in the ordered pairs.

Solution Set
 {(1,-5),(5,-1)}
CHECK:

    x2 + y2 = 26
    x - y = 6

 

Check:  (1,-5)  - plug into the two equations
(1)2 + (-5)2 = 26
1 + 25 = 26 
26 = 26  Check		 (1) - (-5) = 6
1 + 5 = 6
6 = 6  Check
 

Check:   (5, -1)  - plug into the two equations
(5)2 + (-1)2 = 26
25 + 1 = 26
26 = 26  Check		 (5) - (-1) = 6
5 + 1 = 6
6 = 6  Check