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 Math | Elementary Algebra |
| ACT Mathematics | Elementary Algebra |
| Algebra | Solve systems of equations |
| Elementary Algebra | Evaluation of algebraic expressions through substitution Linear equations |
| Foreign Language | Arabic Subtitled Chinese Subtitled Korean Subtitled Spanish Subtitled |
| Language | English Language |
| One-Variable Equations and Inequalities | Solving Equations |
| Product Type | ACT Math |
| Systems of Equations | Linear and Nonlinear Equations |
Transcript
We get y equals 20 minus 6x.
Now we just substitute 20 minus 6x anywhere we see a y.
Ok so rewrite the above equation with that substitution and we have
3x plus 40 minus 12x equals 13
Simplify to get negative 9x equals 13 minus 40 or negative 9x equals negative 27; divide
both sides by negative 9 and we get x equals 3.
Phew. Great. Only A and B are possible answers.
Now just plug in 3 for x in the second equation
and we get 6 times 3 is 18... plus y... equals 20.
Now subtract 18 from both sides and we get y equals 2.
Here we go. So boom! The answer is B.