ACT Math 2.5 Elementary Algebra

ACT Math Section: Elementary Algebra Drill 2, Problem 5. Use these two equations to solve for both x and y.

ACT MathElementary Algebra
ACT MathematicsElementary Algebra
AlgebraSolve systems of equations
Elementary AlgebraEvaluation of algebraic expressions through substitution
Linear equations
Foreign LanguageArabic Subtitled
Chinese Subtitled
Korean Subtitled
Spanish Subtitled
LanguageEnglish Language
One-Variable Equations and InequalitiesSolving Equations
Product TypeACT Math
Systems of EquationsLinear and Nonlinear Equations

Transcript

00:27

We get y equals 20 minus 6x.

00:30

Now we just substitute 20 minus 6x anywhere we see a y.

00:34

Ok so rewrite the above equation with that substitution and we have

00:37

3x plus 40 minus 12x equals 13

00:41

Simplify to get negative 9x equals 13 minus 40 or negative 9x equals negative 27; divide

00:48

both sides by negative 9 and we get x equals 3.

00:52

Phew. Great. Only A and B are possible answers.

00:55

Now just plug in 3 for x in the second equation

00:58

and we get 6 times 3 is 18... plus y... equals 20.

01:02

Now subtract 18 from both sides and we get y equals 2.

01:06

Here we go. So boom! The answer is B.