Use Substitution to solve 4a + 3c = 40 and 2a + 6c = 38
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Use the substitution method to solve:
4a + 3c = 40
2a + 6c = 38
Check Format
Equation 1 is in the correct format.
Check Format
Equation 2 is in the correct format.
Rearrange Equation 2 to solve for a:
2a + 6c = 38
Subtract 6c from both sides to isolate a:
2a + 6c - 6c = 38 - 6c
2a = 38 - 6c
Now divide by 2:
2a
2
=
38 - 6c
2
Revised Equation 2:
a =
38 - 6c
2
Plug Revised Equation 2 value into a:
4(a) + 3c = 40
4 * ((38 - 6c)/2) + 3c = 40
((152 - 24c)/2) + 3c = 40
Multiply equation 1 through by 2
2 * (((152 - 24c)/2) + 3c = 40)
2 * (((152 - 24c)/2) + 3c = 40)
152 - 24c + 6c = 80
Group like terms:
-24c + 6c = 80 - 152
-18c = -72
Divide each side by -18
-18c
-18
=
-72
-18
c =
-72
-18
c = 4
Plug this answer into Equation 1
4a + 3(4) = 40
4a + 12 = 40
4a = 40 - 12
4a = 28
Divide each side by 4
4a
4
=
28
4
a =
28
4
a = 7
What is the Answer?
a = 7 and c = 4
How does the Simultaneous Equations Calculator work?
Free Simultaneous Equations Calculator - Solves a system of simultaneous equations with 2 unknowns using the following 3 methods: 1) Substitution Method (Direct Substitution) 2) Elimination Method 3) Cramers Method or Cramers Rule
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