# 6. Algebraic Solution of Systems of Equations

## Solution by Substitution

Similar to the linear case in the previous section, we can solve a system of equations by substitutiong one of the expressions given into the other expression. Our solution will be a set of `x`-`y` coordinates.

### Example 1

Solve the system of equations algebraically:

y=x+ 1

x^{2}+y^{2}= 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

x^{2}+y= 5

x^{2}+y^{2}= 25

### Exercises

1. Solve algebraically:

6

y−x= 6

x^{2}+ 3y^{2}= 36

2. Solve algebraically:

3

x^{2}−y^{2}= 4

x^{2}+ 4y^{2}= 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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