Solving Systems of Equations Algebraically

1. When you graph, sometimes you cannot find the exact point of intersection. We can use algebra to find the exact point.

2. Also, we do not need to put every equation in slope-intercept form in order to determine if the lines are parallel or the same line. Algebraic methods will give us the same information.

Methods of Solving Systems Algebraically

We will look at TWO methods to solve systems algebraically:

1) Substitution

2) Elimination

Method 1: SubstitutionSteps:

1. Choose one of the two equations and isolate one of the variables.

2. Substitute the new expression into the other equation for the variable.

3. Solve for the remaining variable.

4. Substitute the solution into the other equation to get the solution to the second variable.

Method 1: Substitution

Example:

Equation ‘a’: 3x + 4y = - 4

Equation ‘b’: x + 2y = 2

Isolate the ‘x’ in equation ‘b’:

x = - 2y + 2

Method 1: Substitution

Example, continued:

Equation ‘a’: 3x + 4y = - 4

Equation ‘b’: x + 2y = 2

Substitute the new expression,

x = - 2y + 2 for x into equation ‘a’:

3(- 2y + 2) + 4y = - 4

Method 1: Substitution

Example, continued:

Equation ‘a’: 3x + 4y = - 4

Equation ‘b’: x + 2y = 2

Solve the new equation:

3(- 2y + 2) + 4y = - 4

- 6y + 6 + 4y = - 4

- 2y + 6 = - 4

- 2y = - 10

y = 5

Method 1: SubstitutionExample, continued:

Equation ‘a’: 3x + 4y = - 4

Equation ‘b’: x + 2y = 2

Substitute y = 5 into either equation ‘a’ or ‘b’:

x + 2 (5) = 2

x + 10 = 2

x = - 8

The solution is (-8, 5).

Method 2: EliminationSteps: 1. Line up the two equations using

standard form (Ax + By = C).

2. GOAL: The coefficients of the same variable in both equations should have the same value but opposite signs.

3. If this doesn’t exist, multiply one or both of the equations by a number that will make the same variable coefficients opposite values.

Method 2: Elimination

Steps, continued:

4. Add the two equations (like terms).

5. The variable with opposite coefficients should be eliminated.

6. Solve for the remaining variable.

7. Substitute that solution into either of the two equations to solve for the other variable.

Method 2: Elimination

Example:

Equation ‘a’: 2x - 4y = 13Equation ‘b’: 4x - 5y = 8

Multiply equation ‘a’ by –2 to eliminate the x’s:

Equation ‘a’: -2(2x - 4y = 13)Equation ‘b’: 4x - 5y = 8

Method 2: EliminationExample, continued:

Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8

Add the equations (the x’s are eliminated):

-4x + 8y = -26 4x - 5y = 8 3y = -18 y = -6

Method 2: Elimination

-224

-11

2

Example, continued:

Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8

Substitute y = -6 into either equation:

Solution: ( , -6)-11

2

4x - 5(-6) = 84x + 30 = 8

4x = -22x = x =

Method 2: Elimination

Example 2: Equation ‘a’: -9x + 6y = 0Equation ‘b’: -12x + 8y = 0

Multiply equation ‘a’ by –4 and equation ‘b’ by 3 to eliminate the x’s:

Equation ‘a’: - 4(-9x + 6y = 0)Equation ‘b’: 3(-12x + 8y = 0)

36x - 24y = 0

-36x + 24y = 0

0 = 0

Method 2: Elimination

Example 2, continued:

Equation ‘a’: - 4(-9x + 6y = 0)Equation ‘b’: 3(-12x + 8y = 0)

What does this answer mean?

Is it true?

36x - 24y = 0

-36x + 24y = 0

0 = 0

Method 2: Elimination

Example 2, continued:

When both variables are eliminated, if the statement is TRUE (like 0 = 0),

then they are the same lines and there are infinite solutions.

if the statement is FALSE (like 0 = 1), then they are parallel lines and there is no solution.

36x - 24y = 0

-36x + 24y = 0

0 = 0

Method 2: Elimination

Example 2, continued:

Since 0 = 0 is TRUE, there are infinite solutions.

Solving Systems of Three Equations Algebraically

1. When we have three equations in a system, we can use the same two methods to solve them algebraically as with two equations.

2. Whether you use substitution or elimination, you should begin by numbering the equations!

Solving Systems of Three Equations

Substitution Method 1. Choose one of the three equations

and isolate one of the variables.

2. Substitute the new expression into each of the other two equations.

3. These two equations now have the same two variables. Solve this 2 x 2 system as before.

4. Find the third variable by substituting the two known values into any equation.

Solving Systems of Three Equations

Linear Combination Method1. Choose two of the equations and eliminate

one variable as before.

2. Now choose one of the equations from step 1 and the other equation you didn’t use and eliminate the same variable.

3. You should now have two equations (one from step 1 and one from step 2) that you can solve by elimination.

4. Find the third variable by substituting the two known values into any equation.

Systems of Nonlinear Equations and Their Solutions

A system of two nonlinear equations in two variables contains at least one equation that cannot be expressed in the form Ax + By = C. Here are two examples:

A solution to a nonlinear system in two variables is an ordered pair of real numbers that satisfies all equations in the system. The solution set to the system is the set of all such ordered pairs.

x2 = 2y + 103x – y = 9

y = x2 + 3x2 + y2 = 9

Example: Solving a Nonlinear System by the Substitution Method

Solve by the substitution method:The graph is a line.

The graph is a circle.

x – y = 3 (x – 2)2 + (y + 3)2 = 4

Solution Graphically, we are finding the intersection of a line and a circle whose center is at (2, -3) and whose radius measures 2.

Step 1 Solve one of the equations for one variable in terms of the other. We will solve for x in the linear equation - that is, the first equation. (We could also solve for y.)

x – y = 3 This is the first equation in the given system.

x = y + 3 Add y to both sides.

Solution

Step 2 Substitute the expression from step 1 into the other equation. We substitute y + 3 for x in the second equation.

x = y + 3 ( x – 2)2 + (y + 3)2 = 4

This gives an equation in one variable, namely

(y + 3 – 2)2 + (y + 3)2 = 4.

The variable x has been eliminated.

Step 3 Solve the resulting equation containing one variable.(y + 3 – 2)2 + (y + 3)2 = 4 This is the equation containing one variable.

(y + 1)2 + (y + 3 )2 = 4 Combine numerical terms in the first parentheses.

y2 + 2y + 1 + y2 + 6y + 9 = 4 Square each binomial.

2y2 + 8y + 10 = 4 Combine like terms on the left.

2y2 + 8y + 6 = 0 Subtract 4 from both sides and set the quadratic equation equal to 0.

Solution

y2 + 4y + 3 = 0 Simplify by dividing both sides by 2.

(y + 3)(y + 1) = 0 Factor.

y + 3 = 0 or y + 1 = 0 Set each factor equal to 0.

y = -3 or y = -1 Solve for y.

-1-5 -4 -3 -2 1 2 3 4 5 6 7

5

4

3

2

1

7

6

-3

-4

-5

-6

-7

-1

-2

x – y = 3

(x – 2)2 + (y + 3)2 = 4

(2, -1)

(0, -3)

Step 4 Back-substitute the obtained values into the equation from step 1. Now that we have the y-coordin-ates of the solutions, we back-substitute -3 for y and -1 for y in the equation x = y + 3.

If y = -3: x = -3 + 3 = 0, so (0, -3) is a solution.If y = -1: x = -1 + 3 = 2, so (2, -1) is a solution.

Step 5 Check the proposed solution in both of the system's given equations. Take a moment to show that each ordered pair satisfies both equations. The solution set of the given system is {(0, -3), (2, -1)}.

Example: Solving a Nonlinear System by the Addition Method

Solve the system:Equation 1.

Equation 2.

4x2 + y2 = 13 x2 + y2 = 10

Solution We can use the same steps that we did when we solved linear systems by the addition method.

Step 1 Write both equations in the form Ax2 + By2 = C. Both equations are already in this form, so we can skip this step.

Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x2-coefficients or the sum of the y2-coefficients is 0. We can eliminate y by multiplying Equation 2 by -1.

No change.

Multiply by -1.4x2 + y2 = 13x2 + y2 = 10

4x2 + y2 = 13-x2 – y2 = -10

SolutionSteps 3 and 4 Add equations and solve for the remaining variable.

Add.

Step 5 Back-substitute and find the values for the other variables. We must back-substitute each value of x into either one of the original equations. Let's use x2 + y2 = 10, Equation 2. If x = 1, 12 + y2 = 10 Replace x with 1 in Equation 2.

y2 = 9 Subtract 1 from both sides.

y = ±3 Apply the square root method.

(1, 3) and (1, -3) are solutions. If x = -1, (-1)2 + y2 = 10 Replace x with -1 in Equation 2.

y2 = 9 The steps are the same as before. y = ±3 (-1, 3) and (-1, -3) are solutions.

2 2

2 2

2

2

4 13

- - -10

3 3

1

1

x y

x y

x

x

x

-1-5 -4 -3 -2 1 2 3 4 5 6 7

54

3

2

1

7

6

-3

-4

-5

-6

-7

-1

-2

4x2 + y2 = 13

x2 + y2 = 10

(-1, -3)

(-1, 3) (1, 3)

(1, -3)

Step 6 Check. Take a moment to show that each of the four ordered pairs satisfies Equation 1 and Equation 2. The solution set of the given system is{(1, 3), (1, -3), (-1, 3), (-1, -3)}.

Solution

Example: Solving a Nonlinear System by the Addition Method

Solve the system: y = x2 + 3 Equation 1 (The graph is a parabola.)

x2 + y2 = 9 Equation 2 (The graph is a circle.)

Solution We could use substitution because Equation 1 has y expressed in terms of x, but this would result in a fourth-degree equation. However, we can rewrite Equation 1 by subtracting x2 from both sides and adding the equations to eliminate the x2-terms.

Add.

-x2 + y = 3x2 + y2 = 9

y + y2 = 12

Subtract x2 from both sides of Equation 1.This is Equation 2.Add the equations.

Solution

We now solve this quadratic equation. y + y2 = 12y2 + y – 12 = 0 Subtract 12 from both circles and get the quadratic

equation equal to 0.

(y + 4)(y – 3) = 0 Factor.

y + 4 = 0 or y – 3 = 0 Set each factor equal to 0.

y = -4 or y = 3 Solve for y.

To complete the solution, we must back-substitute each value of y into eitherone of the original equations. We will use y = x2 + 3, Equation 1. First, wesubstitute -4 for y.

-4 = x2 + 3-7 = x2 Subtract 3 from both sides.

-1-5 -4 -3 -2 1 2 3 4 5 6 7

54

3

2

1

7

6

-3

-4

-5

-6

-7

-1

-2

y = x2 + 3

x2 + y2 = 9

(0, 3)

Because the square of a real number cannot be negative, the equation x2 = -7 does not have real-number solutions. Thus, we move on to our other value for y, 3, and substitute this value into Equation 1.

Solution

y = x2 + 3 This is Equation 1.

3 = x2 + 3 Back-substitute 3 for y.

0 = x2 Subtract 3 from both sides.

0 = x Solve for x.

We showed that if y = 3, then x = 0. Thus, (0, 3) is the solution. Take a moment to show that (0, 3) satisfies Equation 1 and Equation 2. The solution set of the given system is {(0, 3)}.

Examples

Solve:

4. Find the length and width of a rectangle whose perimeter is 20 ft. an whose area is 21 sq.ft.

2

2 2 2 2

2 2

2 2

2 0 51. 3.

1 1 5 25

3 2 352.

4 3 48

x y y x

x y x y

x y

x y