Algebraic Manipulation Worksheets With Answers - Craven Community College Page 23
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29Substitute Y = 1 into the first equation and solve for X.
2X – 5Y = 1
First equation
2X – 5(1) = 1
Substitute for Y and Multiply
2X – 5 = 1
Add 5 to both sides
2X = 6
Divide by 2
X = 3
Check:
2(3) – 5(1) = 1
3(3) – 4(1) = 5
Thus the solution to this system is X = 3 and Y = 1
Solving Systems of Equations by Substitution
Another common algebraic method for solving systems of equations is the Substitution Method. In this
method one equation is solved for one of the variables. This algebraic expression is then substituted into
the other equation, producing a new equation in a single variable. This new equation is solved and the
resulting value substituted back into an original equation to solve for the other variable
Example:
Y = 3X – 7
4Y – 3X = 8
Since the first equation is already in the form solving for Y, 3X – 7 will be substituted for the Y in
the second equation.
4Y – 3X = 8
Second equation
4(3X – 7) = 8
Substitute 3X — 7 for Y
12X – 28 – 3X = 8
Multiply & combine like terms
9X – 28 = 8
Add 28 to both sides
9X = 36
Divide by 9
X = 4
Now that a value for X has been found, substitute this value back into the first original equation to
obtain a value for the Y variable.
Y = 3X – 7
First equation
Y = 3(4) – 7
Substitute 4 for X
Y = 12 – 7
Multiply combine like terms
Y = 5
Check:
5 = 3(4) – 7
4(5) – 3(4) = 8
Thus the solution to this system is X = 4 and Y = 5.
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