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
x2 + y2 = 25
Answer
We recognize that this is a straight line intersecting a circle. (See more on the circle.)
We may have:
- no intersection point
- 1 intersection point
- 2 intersection points
We can simply substitute the right hand side of the first equation into the second equation:
x2 + (x + 1)2 = 25
This gives:
x2 + x2 + 2x + 1 = 25
2x2 + 2x − 24 = 0
x2 + x − 12 = 0
(x + 4)(x − 3) = 0
So `x = −4` or `x = 3`.
This gives our intersecting points to be: `(−4, −3)` and `(3, 4)`.
Is it correct? The graph showing the line intersecting the circle is as follows:
Graphs of `y = x+1` and `x^2+y^2=25`: Intersection line and circle
We can see from the graph that our solutions `(−4, −3)` and `(3, 4)` are correct.
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
Answer
NOTE: This system represents a parabola intersecting a circle. We expect:
- no intersection point or possibly
- 1, 2, 3 or 4 intersection points
If we subtract the first line from the second, we have:
y2 − y = 25 − 5
y2 − y − 20 = 0
(y + 4)(y − 5) = 0
So y = −4 or 5
The corresponding x values are going to be:
`x = +3` or ` −3`, and `x = 0`
So the solution set will be:
`(−3,−4)`, `(3,−4)` and `(0,5)`.
The sketch shows:
Graphs of `y = -x^2+5` and `x^2+y^2=25`: Intersection parabola and circle
We can see from the graph that our 3 solutions `(−3,−4)`, `(3,−4)` and `(0, 5)` are correct.
Exercises
1. Solve algebraically:
6y − x = 6
x2 + 3y2 = 36
Answer
We first solve the first line for y (so we can substitute):
`y=(x+6)/6`
Substituting in the second row gives:
`x^2+3((x+6)/6)^2=36`
Expand the brackets:
`x^2+3((x^2+12x+36)/36)=36`
Cancel the 3 and 36:
`x^2+((x^2+12x+36)/12)=36`
Multiply throughout by 12:
`12x^2+x^2+12x+36=432`
`13x^2+12x-396=0`
Solving using the quadratic formula gives:
`x = 5.077`, or `x = −6`
This gives us solutions of: `(5.077, 1.85)` and `(−6,0)`.
Graphically, we have:
Graphs of `y = (x+6)/6` and `x^2+3y^2=36`: Intersection of line and ellipse
2. Solve algebraically:
3x2 − y2 = 4
x2 + 4y2 = 10
Answer
The first equation is a hyperbola, while the second is an ellipse.
We multiply the first row by 4 so we can eliminate the `y^2` term:
`12x^2-4y^2=16`
`x^2+4y^2=10`
Now adding the two rows, we obtain:
`13x^2=26`
`x^2=2`
`x=+-sqrt(2)`
Substituting `+sqrt(2)` into the question's first equation gives us
`y=+-sqrt(2)`
Likewise, substituting `-sqrt(2)` into the first equation also gives us
`y=+-sqrt(2)`
This gives us the solutions
`(sqrt(2),sqrt(2)),` `(sqrt(2),-sqrt(2)),` `(-sqrt(2),sqrt(2)),` `(-sqrt(2),-sqrt(2)) `
≈ (±1.414, ±1.414), (±1.414, ∓1.414)
The graph shows the intersection of the ellipse and the hyperbola. We see 4 intersection points, with the same values that we found algebraically.
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.
Answer
The statement "R is numerically equal to the square of the reactance X" simply means `R = X^2`.
Recall (from Application of Complex Numbers) that
`|Z|=sqrt(R^2+(X_L-X_C)^2`
In this case, from the definition, and to make life easier, we assume that XL − XC = X.
So
`|Z|=sqrt(R^2+X^2)=2Omega`
Now, on squaring both sides, we have
R2 + X2 = 4
But R = X2 (since R is equal to the square of X) so
R2 + R = 4
Then
R2 + R − 4 = 0
Using quadratic formula gives
`R=(-1+-sqrt(1+16))/2` `=(-1+-sqrt17)/2`
Only the positive root has meaning (since we cannot have negative resistance), so
R = 1.56 Ω. and therefore X = √1.56 = 1.25 Ω.
In this question, the 2 interssecting functions are a parabola (`X = R^2 + R - 4`) and a straight line (`X=4`). In the graph, we can see the 2 solutions we obtained, one is negative (`R=-2.56`) and the other one is positive, at (`R=1.56`).
4. Find the intersection points for the circles
(x + 2)2 + (y − 3)2 = 25
and
(x − 1)2 + (y + 4)2 = 16
Answer
a. Setting the right side to 0 and expanding each equation gives:
Equation [1]:
(x + 2)2 + (y − 3)2 = 25
(x + 2)2 + (y − 3)2 − 25 = 0
x2 + 4x + 4 + y2 − 6y + 9 − 25 = 0
x2 + 4x + y2 − 6y − 12 = 0 [3]
Equation [2]:
(x − 1)2 + (y + 4)2 = 16
(x − 1)2 + (y + 4)2 − 16 = 0
x2 − 2x + 1 + y2 + 8y + 16 − 16 = 0
x2 − 2x + y2 + 8y + 1 = 0 [4]
Solving the above 2 results simultaneously, we subtract Row [4] from Row [3] and this gives:
6x − 14y − 13 = 0
Solving for y gives:
`y=3/7x-13/14`
This means the intersection points are on the line
`y=3/7x-13/14`
b. Solve one of the circle equations for y using:
`y=(-b+-sqrt(b^2-4ac))/(2a)`
`y^2-6y+(x^2+4x-12)=0`
This gives:
`y=3+-sqrt(21-x^2-4x`
c. Substitute the positive case into the LHS of
`y=3/7x-13/14`, gives us:
`3+sqrt(21-x^2-4x)=3/7x-13/14`
d. Solve for x:
`sqrt(21-x^2-4x)` `=3/7x-13/14-3` `=3/7x-55/14`
Square both sides:
`21-x^2-4x` ` =(3/7x-55/14)^2` `=3025/196-165/49x+9/49x^2`
Re-writing for convenience:
`21-x^2-4x` `=3025/196-165/49x+9/49x^2`
Moving everything to the right hand side:
`0=9/49x^2+x^2-165/49x` `+4x` `+3025/196` `-21`
Simplifying:
`0=58/49x^2+31/49x-1091/196`
This gives us a quadratic in x
`58/49x^2+31/49x-1091/196=0`
or more simply
`58x^2+31x-272 3/4=0`
`x=-31/116+7/116sqrt1311` `=1.9177`
`x=-31/116-7/116sqrt1311` `=-2.4522`
e. Substitute these two x-values into
`y=3/7x-13/14`:
`[3/7x-13/14]_(x=1.9177)=-0.1067`
`[3/7x-13/14]_(x=-2.4522)=-1.9795`
So the points of intersection are (1.9177, −0.1067) and (−2.4522, −1.9795).
[We could have substituted the x-values into either circle equation and solved for y, but what I have done is easier.]
We can see that the circles, the line `y=3/7x-13/14` and the intersection points are all correct when we draw the graph:
As a comment, this question could be solved very quickly, and easily, using a computer graphics program. We would just need to zoom in on the intersection points until we obtained the required precision.