Pennsylvania Algebra 1 - 2020 Edition
5.02 Solving systems of equations with the substitution method
Lesson

Recall that we can solve systems of equations graphically. Now, we'll look at how to solve a system of equations algebraically, using the substitution method. This method works the same way as the substitution property; which states: if $a=b$a=b then $b$b can be replaced with $a$a in any equation or inequality. Therefore, solve an equation from the system for one variable (as you would for a literal equation) then substitute the value (expression) into the other equation.

#### Worked example

##### Example 1

Solve the following system using the substitution method.

eq 1: $3y-x=-4$3yx=4 and eq 2: $2+x-y=0$2+xy=0

Think: To solve the system, we'll first pick one literal equation to solve for either $x$x or $y$y .

Do: The second equation (eq 2) looks easier to manipulate, and transforming it gives us $y=2+x$y=2+x. Knowing the exact value of $y$y means we can combine the two equations by substituting $y=2+x$y=2+x into $3y-x=-4$3yx=4(eq 1). This gives us:

 $3\left(2+x\right)-x$3(2+x)−x $=$= $-4$−4 $6+3x-x$6+3x−x $=$= $-4$−4 $6+2x$6+2x $=$= $-4$−4 $2x$2x $=$= $-4-6$−4−6 $2x$2x $=$= $-10$−10 $x$x $=$= $-5$−5

Now that we know the value of $x$x we can easily solve for $y$y by using the equation that was not substituted into: $2+x-y=0$2+xy=0 or $y=2+x$y=2+x (eq 2). Therefore, $y=2+\left(-5\right)$y=2+(5) which equals $-3$3. The point of intersection (solution) answer is $(-5,-3)$(5,3).

Reflect: Check your answer. Remember, when you substitute the $x$x- and $y$y- values into the equations they must create true statements in order for them to be the solution.

eq 1 eq 2
$3y-x=-4$3yx=4 $2+x-y=0$2+xy=0
$3\times\left(-3\right)-\left(-5\right)=-4$3×(3)(5)=4 $2+\left(-5\right)-\left(-3\right)=0$2+(5)(3)=0
$-9+5=-4$9+5=4 $-3+3=0$3+3=0
$-4=-4$4=4 $0=0$0=0

Tips!
• Being organized is key. Name your equations by writing $(1)$(1) & $(2)$(2) next to them, and whenever you create new equations out of one or both of them, you can name them $(3)$(3), $(4)$(4), etc. When substituting equation $(3)$(3) into equation $(1)$(1) you can write it in shorthand as $(3)$(3)$(1)$(1). This helps to keep your ideas in order and not get confused.
• When dealing with equations with big numbers, see if you can simplify them before beginning to solve them, eg. $2x-4=6y$2x4=6y can be simplified to $x-2=3y$x2=3y without changing the values of $x$x & $y$y.
Careful!

Remember to solve for the values of both $x$x and $y$y. Check if you have used both at the end of every system of equations problem unless told otherwise.

### Concurrent lines

We know that the solution of a system of equations is represented graphically as the point of intersection for the two lines, or where they cross over. But did you know that system of equations can also tell us whether a set of three or more lines are concurrent? When a set of lines are concurrent it means that they all cross over at the same point, like in the diagram below:

Having the same intersection point between them means that if you take any two of these three lines then their intersection point $P$P will be the same as that of any other two lines.

#### Worked example

##### Example 2

Are the lines below concurrent?

 Equation 1 $-13y-2=5x$−13y−2=5x Equation 2 $8=5y-x$8=5y−x Equation 3 $y-3x=3$y−3x=3

(a) First solve Equations 1 & 2 using the substitution method

Think: First look for an $x$x or $y$y term in either equation that is easy to isolate and make it the subject

Do: The $x$x term in Equation 2 is easy to isolate:

 $8$8 $=$= $5y-x$5y−x $8+x$8+x $=$= $5y$5y $x$x $=$= $5y-8$5y−8

Substituting this expression for $x$x into Equation 1:

 $-13y-2$−13y−2 $=$= $5x$5x $=$= $5\left(5y-8\right)$5(5y−8) $=$= $25y-40$25y−40 $40-2$40−2 $=$= $25y+13y$25y+13y $38$38 $=$= $38y$38y $y$y $=$= $1$1

Substituting this $y$y value back into $x=5y-8$x=5y8:

 $x$x $=$= $5y-8$5y−8 $=$= $5\times1-8$5×1−8 $=$= $5-8$5−8 $=$= $-3$−3

So the solution to Equations 1 & 2 is $(-3,1)$(3,1)

(b) Solve Equations 2 & 3 using the substitution method

Think: First look for an $x$x or $y$y term in either equation that is easy to isolate and make it the subject

Do: Let's use Equation 2 again and make $x$x the subject to become $x=5y-8$x=5y8 which we worked out in part (a).

Now let's substitute it into Equation 3:

 $y-3x$y−3x $=$= $3$3 $y-3\left(5y-8\right)$y−3(5y−8) $=$= $3$3 $y-15y+24$y−15y+24 $=$= $3$3 $-14y$−14y $=$= $-21$−21 $y$y $=$= $\frac{3}{2}$32​

Substituting this $y$y value back into $x=5y-8$x=5y8:

 $x$x $=$= $5y-8$5y−8 $=$= $5\times\frac{3}{2}-8$5×32​−8 $=$= $\frac{15}{2}-8$152​−8 $=$= $\frac{15-16}{2}$15−162​ $=$= $\frac{-1}{2}$−12​

So the solution to Equations 2 & 3 is $($($\frac{-1}{2},\frac{3}{2}$12,32$)$)

(c) Are these three lines concurrent?

Think: Concurrent lines are three or more lines that intersect at the same point. What does an intersection represent in simultaneous equations?

Do: If the three lines are concurrent, then the intersection point between Equations 1 & 2 must be the same as Equations 2 & 3. We know that graphically the intersection point represents the solution between a system of simultaneous equations. The solution to Equations 1 & 2 is $\left(-3,1\right)$(3,1) which is not equal to $\left(-\frac{1}{2},\frac{3}{2}\right)$(12,32), the solution to Equations 2 & 3. Therefore they can not be concurrent.

#### Practice questions

##### Question 1

We want to solve the following system of equations using the substitution method.

 Equation 1 $y=3x-18$y=3x−18 Equation 2 $x-y=10$x−y=10
1. First solve for $x$x.

2. Hence, solve for $y$y.

##### Question 2

We want to solve the following system of equations using the substitution method.

 Equation 1 $y=5x+34$y=5x+34 Equation 2 $y=3x+18$y=3x+18
1. First solve for $x$x.

2. Hence, solve for $y$y.

##### Question 3

We want to determine whether the lines below are concurrent.

 Equation 1 $-6x-2y=-28$−6x−2y=−28 Equation 2 $2x+16y=40$2x+16y=40 Equation 3 $4x-2y=12$4x−2y=12
1. First use the substitution method to find the values of $x$x and $y$y that satisfy Equation $1$1 and $2$2 simultaneously. Enter both solutions on one line, separated by a comma.

 Equation 1 $-6x-2y=-28$−6x−2y=−28 Equation 2 $2x+16y=40$2x+16y=40
2. Now determine if this solution satisfies equation 3 by substituting the values of $x$x and $y$y into the left hand side:

 Equation 3 $4x-2y=12$4x−2y=12
3. Are the lines represented by Equations 1, 2 and 3 concurrent?

No

A

Not enough information.

B

Yes

C

No

A

Not enough information.

B

Yes

C

### Outcomes

#### A1.1.2.2.1

Write and/or solve a system of linear equations (including problem situations) using graphing, substitution, and/or elimination. Note: Limit systems to two linear equations.