Transcript Solving Systems with Substitution

Dr. Fowler  CCM
Solving Systems of Equations
By Substitution – Harder
Solving a system of equations by substitution
Step 1: Solve an equation
for one variable.
Pick the easier equation. The goal
is to get y= ; x= ; a= ; etc.
Step 2: Substitute
Put the equation solved in Step 1
into the other equation.
Step 3: Solve the equation.
Get the variable by itself.
Step 4: Plug back in to find
the other variable.
Substitute the value of the variable
into the equation.
Step 5: Check your
solution.
Substitute your ordered pair into
BOTH equations.
ALREADY IN NOTES – Read Only for Review
1) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 1: Solve an equation
for one variable.
Step 2: Substitute
It is easiest to solve the
first equation for x.
3y + x = 7
-3y
-3y
x = -3y + 7
4x – 2y = 0
4(-3y + 7) – 2y = 0
1) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 3: Solve the equation.
-12y + 28 – 2y = 0
-14y + 28 = 0
-14y = -28
y=2
Step 4: Plug back in to find
the other variable.
4x – 2y = 0
4x – 2(2) = 0
4x – 4 = 0
4x = 4
x=1
1) Solve the system using substitution
3y + x = 7
4x – 2y = 0
(1, 2)
3(2) + (1) = 7
4(1) – 2(2) = 0
Step 5: Check your
solution.
Answer is
(1,2)
2) Solve the following system using the substitution method.
3x – y = 6
and – 4x + 2y = –8
STEP 1 – Solve the first equation for y is easiest,
3x – y = 6
–y = –3x + 6 (subtract 3x from both sides)
y = 3x – 6
(multiply both sides by – 1)
STEP 2 – Substitute this value for y in the OTHER equation.
–4x + 2y = –8
–4x + 2(3x – 6) = –8 (replace y to other equation)
–4x + 6x – 12 = –8 (use the distributive property)
2x – 12 = –8 (simplify the left side)
2x = 4
(add 12 to both sides)
x=2
(divide both sides by 2)
CONTINUED >
STEP 3 – To get y, substitute x = 2 into either original
equation. The easiest is the one that was already
solved for y.
y = 3x – 6
= 3(2) – 6 = 6 – 6 = 0
y=0
We have now found x & y.
Answer is
(2, 0)
EXAMPLE 3
Solve the system by the substitution method.
x  3 y  7
4 x  12 y  28
Solve first for x:
x  3 y  3 y  7  3 y
x  7  3 y
TRUE – The answer is
Substitute:
4  7  3 y   12 y  28
28  12 y  12 y  28
28  28
infinitely many solutions
EXAMPLE 4
Use substitution to solve the system.
Solve first for x:
x  1  1  4 y  1
x  4 y  1
x  1  4 y
2 x  5 y  11
Substitute into other equation:
2  4 y 1  5 y  11
8 y  2  5 y  2  11  2
13 y 13
To get X, substitute y you found into

equation already solved for X:
13 13
x  4  1  1
y  1
x  4 1
 3, 1
x3
Example #5:
x + y = 10
5x – y = 2
Step 1: Solve one equation for one variable.
x + y = 10
y = -x +10
Step 2: Substitute into the other equation.
5x - y = 2
5x -(-x +10) = 2
x + y = 10
5x – y = 2
Step 3: Simplify and solve the equation.
5x -(-x + 10) = 2
5x + x -10 = 2
6x -10 = 2
6x = 12
x=2
x + y = 10
5x – y = 2
Step 4: Substitute back into either original
equation to find the value of the other
variable.
x + y = 10
2 + y = 10
y=8
Solution to the system is (2,8).
Excellent Job !!!
Well Done
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