6.05 Solving systems of equations with a variety of methods
There are essentially three different methods to solve systems of equations. They are listed and described briefly below.
Graphical Method
It is easy to use a graphing calculator or utility when the equations within a system have y isolated. In this case, the calculator can be used to graph the equations. The intersection of the two lines will represent the solution to the system of equations.
Worked examples
Question 1
Solve the following system of equations:
$y=5x$y=5x
$y=x+2$y=x+2
Solve:
Here I have drawn the $y=5x$y=5x line as red and the $y=x+2$y=x+2 line as green:
We can then see that there is only one intersection point and it is $(0.5,2.5)$(0.5,2.5). Therefore the solution that solves the two equations must be when $x=0.5$x=0.5 and $y=2.5$y=2.5.
Substitution Method
There are two different types of systems of equations where substitution is the easiest method.
Type 1: One variable is isolated in one of the equations. The system is solved by substituting the equation with the isolated term into the other equation.
Worked examples
question 2
Solve following system of equations:
$3y+4=x$3y+4=x
$2+x-y=0$2+x−y=0
Solve:
| $2+3y+4-y$2+3y+4−y | $=$= | $0$0 |
| $2+4+3y-y$2+4+3y−y | $=$= | $0$0 |
| $6+2y$6+2y | $=$= | $0$0 |
| $2y$2y | $=$= | $-6$−6 |
| $y$y | $=$= | $-3$−3 |
Now that we know the value of $y$y we can easily solve for $x$x by using the first equation: $3y+4=x$3y+4=x. Therefore $3\times\left(-3\right)+4=x$3×(−3)+4=x which equals $-5$−5. So our answer is $(-5,-3)$(−5,−3).
Type 2: One variable can be easily isolated. Solve for one variable in one of the equations, then substituting that equation into the second equation. Solve for a in the second equation, then substitute the second equation into the first.
Worked example
question 3
Solve the following system of equations:
$2p-3q=5$2p−3q=5
$p-2q=8$p−2q=8
Solve:
Isolate $p$p in equation 2 - $p=2q+8$p=2q+8
Substitute the value of p into equation 1.
| $2\left(2q+8\right)-3q$2(2q+8)−3q | $=$= | $5$5 |
| $4q+16-3q$4q+16−3q | $=$= | $5$5 |
| $q+16$q+16 | $=$= | $5$5 |
| $q$q | $=$= | $-11$−11 |
Now that we know the value of $q$q we can easily solve for $p$p by using the second equation (modified version): $p=2q+8$p=2q+8 . Therefore$p=2\times\left(-11\right)+8$p=2×(−11)+8 which equals $-14$−14. The answer is $(-14,-11)$(−14,−11).
Elimination Method
When both equations within a system are given in standard form: Ax + By = C. The system of equations are usually solved by eliminating a variable and solving for the remaining variable.
Worked example
Question 4
Solve the following system of equations:
$2x-y=1$2x−y=1
$5x+y=2$5x+y=2
Solve:
| $2x-y+5x+y$2x−y+5x+y | $=$= | $1+2$1+2 |
| $2x+5x-y+y$2x+5x−y+y | $=$= | $3$3 |
| $7x$7x | $=$= | $3$3 |
| $x$x | $=$= | $\frac{3}{7}$37 |
Now, find the y value by substituting the above x value into the either equations.
| $5\times\frac{3}{7}+y$5×37+y | $=$= | $2$2 |
| $\frac{15}{7}+y$157+y | $=$= | $2$2 |
| $y$y | $=$= |
$\frac{14}{7}-\frac{15}{7}$147−157 |
| $y$y | $=$= | $\frac{-1}{7}$−17 |
Solution: $\text{(-3/7, -1/7)}$(-3/7, -1/7)
We can see that we should use addition when the coefficients of the variable we want to eliminate are equal and opposite in sign, and subtraction when they're equal and the same sign. But what happens when they don't have the same coefficients at all?
question 5
Solve the following system of equations:
$x+3y=5$x+3y=5
$2y+2x=1$2y+2x=1
Solve:
Multiplying the first equation by $-2$−2:
| $-2\left(x+3y\right)$−2(x+3y) | $=$= | $-2\times5$−2×5 |
| $-2x-6y$−2x−6y | $=$= | $-10$−10 |
Add the modified Equation 1 and Equation 2:
| $-2x-6y+2y+2x$−2x−6y+2y+2x | $=$= | $-10+1$−10+1 |
| $-2x-6y+2y+2x$−2x−6y+2y+2x | $=$= | $-9$−9 |
| $-4y$−4y | $=$= | $-9$−9 |
| $y$y | $=$= | $\frac{9}{4}$94 |
Now that we have our value for $y$y it's a simple case of substituting it back in any of the equations to get our value for $x$x:
| $x+3y$x+3y | $=$= | $5$5 |
| $x+\frac{3\times9}{4}$x+3×94 | $=$= | $5$5 |
| $x+\frac{27}{4}$x+274 | $=$= | $5$5 |
| $x$x | $=$= | $5-\frac{27}{4}$5−274 |
| $x$x | $=$= | $\frac{-7}{4}$−74 |
Solution: $\text{(-7/4, 9/4)}$(-7/4, 9/4)
Practice questions
question 6
Here is a system of linear equations.
Equation 1: $y=x+8$y=x+8
Equation 2: $y=-7x+16$y=−7x+16
Solve for the value of $x$x that satisfies both equations.
Hence determine the value of $y$y that satisfies both equations.
Add equations 1 and 2 to create equation 3.
State equation 3. Give your answer in the form $y=\dots$y=…
Multiply Equation 1 by the number $3$3 to create Equation 4.
State Equation 4.
Solve for the value of $x$x that satisfies the system of equations.
Equation 3: $y=-3x+12$y=−3x+12
Equation 4: $3y=3x+24$3y=3x+24
Are the solutions in the previous part the same as for the system of equations 1 and 2?
Yes, equation 3 is identical to equation 1 and equation 4 is identical to equation 2.
ANo, equations 3 and 4 are a different set of equations to equations 1 and 2, and therefore have different solutions.
BNo, the $y$y-value that satisfies the system of equations 3 and 4 is not equal to the $y$y-value that satifies the system of equations 1 and 2.
CYes, equations 3 and 4 are the result of applying identical operations to both the left hand side and the right hand side of each of the original equations.
D