Algebra I
Solving Systems of Equations Using Substitution
NAME: ______
Today we will be learning how to solve a system of equations using a method known as substitution. Rather than simply listening to me explain how to do it, I’m going to have you and a partner explore and discover the method yourselves. Woo-hoo!
Let’s begin …
EXAMPLE 1
y = x + 1
2x + y = 7
Explain:
In the system of equations above, the first equation is “y = x + 1.”
Explain what it means for two expressions to be equal.
Observe and discover:
y = x + 1
2x + (x + 1) = 7
2x + y= 7
The system of equations is written on the left.
The first step of the substitution method is on the right.
Explain WHAT was done and WHY it would be OK to do that.
Begin to solve:
Using the equation from step 1 of the substitution method (also written below), solve for the variable. (Show your work!)
2x + (x + 1) = 7
x = ______
Finish solving:
y = x + 1
2x + y = 7
Select either equation from the system and write it here: ______
Using your answer from the “Begin to Solve” step above, how could you find the value of y?
Explain in words.
Now actually do the work you just described and find the value of y. (Remember to show your work!)
Let’s see if it would make a difference if we had chosen the other equation from the system to find y.
Write down the equation you didn’t pick … follow the same process … and find the value of y again.
Was your answer the same?
Check:
State the values you found for x and y as an ordered pair (x, y):______
Substitute those values for x and y into both equations and simplify.
y = x + 12x + y = 7
If the result for both equations is a true statement (examples: 5 = 5 … 24= 24 … -2 = -2 … etc.), that means our ordered pair IS a solution to the system because it satisfies (makes true) both equations. Sweet.
Confirm:
Let’s graph the system of equations to confirm our solution from a graphical perspective.
y = x + 1
slope = _____
y-intercept = _____
2x + y = 7
Solve for y to get in slope-intercept form
slope = _____
y-intercept = _____
Graph the 2 equations on the grid above.
State the intersection point: ______
How does this compare to the solution you got using substitution?
Substitution – Show Me What Ya’ Got NAME: ______
(Individual Work)
Explain each step in solving this system of equations using substitution.
y = 4x + 2ORIGINAL SYSTEM OF EQUATIONS
3y – 2x = 26
Explain in words what’s been done in each step.
3(4x + 2) – 2x = 26______
______
______
12x + 6 – 2x = 26______
10x + 6 = 26
- 6 - 6
10x = 20______
10 10______
x = 2
y = 4x + 2 OR3y – 2x = 26______
y = 4(2) + 2 3y – 2(2) = 26______
y = 10 3y – 4 = 26______
+ 4 + 4
3y = 30
3 3
y = 10
SOLUTION:(2, 10)
Now your turn …
Solve these systems of equations using substitution.
(1)y = 4x(2)3x – 2y = 5
3y + 2x = 28x = 15 – y
Substitution – Follow Up Sheet for NotesNAME: ______
Let’s summarize what you’ve learned:
EXAMPLE with work:STEPS:
y = x + 1This is the original problem.
2x + y = 7
Notice in this system of equations at least one of the equations has been solved for one of the variables. (In other words, in at least one of the equations, one of the variables is by itself.) If neither equation has a variable by itself, choose either of the equations and solve it for either of the variables.
y = x + 1
2x + x + 1 = 7(1)Remove one variable from one of the
2x + y= 7equations and replace it with the expression from the other equation.
2x + x + 1 = 7(2)Take the equation from step 1 & solve.
3x + 1 = 7
- 1 -1
3x = 6
3 3
x = 2
y = x + 1OR2x + y = 7(3)Take the answer from step 2, substitute
y = 2 + 1 2(2) + y = 7it into EITHER equation from the system
y = 3 4 + y = 7and solve for the other variable.
- 4 - 4
y = 3
Solution: (2, 3)(4)State your solution as an ordered pair.
y = x + 1AND2x + y = 7(5)Check your solution by substituting the
3 = 2 + 1 2(2) + 3 = 7values you found into BOTH equations.
3 = 3 4 + 3 = 7If both equations result in true
7 = 7statements, you’ve found the solution!
Substitution – Show Me What Ya’ Got (continued)
(Back to partners)
Lastly … let’s shake things up a bit and turn them on their heads. Fun!
(I)SHAKE-UP # 1
The following system of equations is not quite ready to be solved using substitution.
How is this system different compared to the other systems we’ve looked at so far where we HAVE used the substitution method?
y – x = - 2______
x + y = 12______
How could you “fix” one of the equations so that we COULD use substitution easily? Answer in words, and then do so (actually “fix” it). You do not need to solve the system.
______
(II)SHAKE-UP # 2
ORIGINAL SYSTEM
2x + 5y = 16 2x + 5y = 16Finish the problem for me:
2x + y = 4 2x + 5(4 – 2x) = 162x + y = 4
2x + 20 – 10x = 162( ) + y = 4
2x + y = 4 -8x + 20 = 16
-2x - 2x -20 -20
y = 4 – 2x -8x = -4
-8 -8SOLUTION: ______
x = ______
(How is this value of x different from the others we’ve found? Is that OK?)
______
(III)SHAKE-UP # 3
ORIGINAL SYSTEM
y = 4x – 5 8x -2y = 20
8x – 2y = 20 8x - 2(4x – 5) = 20
8x – 8x + 10 = 20
10 = 20
Say What?!
If I solved the equation correctly, and I ended up with a FALSE statement, what must that imply about the solution? If you are unsure, use your graphing calculator to graph the equations. See if that gives you the clue you need.
______
(IV)SHAKE UP # 4
ORIGINAL SYSTEM
y = 2x – 1 3y + 3 = 6x
3y + 3 = 6x 3(2x – 1) + 3 = 6x
6x – 3 + 3 = 6x
6x = 6x
- 6x- 6x
0 = 0
That’s true, but how?!
If I solved the equation correctly, and I ended up with a TRUE statement, what must that imply about the solution? If you are unsure, use your graphing calculator to graph the equations. See if that gives you the clue you need.
______