6. Algebraic Solution of Systems of Equations

Solution by Substitution

Similar to the linear case in the previous section, we can substitute one of the expressions given into the other expression.

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Example 1

Solve the system of equations algebraically:

y = x + 1

x2 + y2 = 25

Solution by Addition or Subtraction

This method works by eliminating one of the variables from the equations. We then find the value(s) of the remaining variable.

Example 2

Solve the system of equations by adding or subtracting

x2 + y = 5

x2 + y2 = 25

Exercises

1. Solve algebraically:

6y − x = 6

x2 + 3y2 = 36

2. Solve algebraically:

3x2 − y2 = 4

x2 + 4y2 = 10

3. An alternating current has impedance given by Z = 2.00 Ω. If the resistance R in the circuit is numerically equal to the square of the reactance X, find R and X.

4. Find the intersection points for the circles

(x + 2)2 + (y − 3)2 = 25

and

(x − 1)2 + (y + 4)2 = 16

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