SOLVING ALGEBRAIC EQUATIONS MATH REVIEW. 1 ST ORDER EQUATIONS 1 ST power on variable (exponent = 1) Only one variable to solve for → there will be one.

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Presentation transcript:

SOLVING ALGEBRAIC EQUATIONS MATH REVIEW

1 ST ORDER EQUATIONS 1 ST power on variable (exponent = 1) Only one variable to solve for → there will be one numerical solution.

EXAMPLE I 6(x – 2) = 4x + 8 6x – 12 = 4x x - 4x 2x – 12 = x = 20 x = 20/2x = 10

EXAMPLE II 7(y – 4) = 2(3 + 3y) 7y – 28 = 6 + 6y -6y - 6y y – 28 = y = 32

RULES for SOLVING EQUATIONS 1.Eliminate all parentheses. Repeat as needed for nested sets of parenthesis. 2.Isolate the variable. Group like terms as necessary. 3.Isolate the constant. 4.Do the inverse of the coefficient. NOTE: Steps 2 and 3 are interchangeable.

EQUATIONS WITH MORE THAN ONE VARIABLE The solution for a single equation will still contain variables. The variable you are solving for needs to be specified.

EXAMPLE I 6(x – a) = 4x + b (solve for x) 6x – 6a = 4x + b - 4x 2x – 6a = b + 6a + 6a 2x = b + 6a x = b/2 + 3a

EXAMPLE II 7(y – a) = 2(b + y) (solve for y) 7y – 7a = 2b + 2y - 2y 5y – 7a = 2b + 7a + 7a 5y = 2b + 7a y = (2b + 7a)/5

Now it is your turn… Try solving these:  4x + 7a = 6 – 5a (solve for x)  2y – b = 4*8 + 3b (solve for y)  F = m∙a (solve for a)(Newton’s 2 nd Law)  D = m/V (solve for V)(Density equation)

A Way to Represent Certain Simple Equations For equations of the form A = B∙C, you may use the following diagram: A B C

Reading your diagram A = BC A B = A/C B C C = A/B

THE END © Lilian Wehner 2012