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:
Example:
Solve the system of equations algebraically:
y = x + 1
x2 + y2 = 25
We recognise that this is a straight line intersecting a circle. We may have:
- no intersection point
- 1 intersection point
- 2 intersection points
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:
Solve the system of equations by adding or subtracting
x2 + y = 5
x2 + y2 = 25
Exercises:
1. Solve algebraically:
2. Solve algebraically:
3. The impedance Z in an alternating-current circuit is 2.00 W. If the resistance R is numerically equal to the square of the reactance X, find R and X.
4. Find the intersection points for the circles
and
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