ACT Math 3.3 Intermediate Algebra
ACT Math: Intermediate Algebra Drill 3, Problem 3. Solve the equation for x.
| ACT Math | Intermediate Algebra |
| ACT Mathematics | Intermediate Algebra |
| Foreign Language | Arabic Subtitled Chinese Subtitled Korean Subtitled Spanish Subtitled |
| Intermediate Algebra | Systems of equations |
| Language | English Language |
| One-Variable Equations and Inequalities | Solving Equations |
| Product Type | ACT Math |
Transcript
It's really just a "substitute n' solve" problem.
In the first equation, y is already isolated, so let's just substitute the first equation
into the second.
4 times the quantity 3 halves x plus 4 equals x plus 1.
Multiplying the 4, we get that 12x over 2... or just 6x... plus 16 equals x plus 1.
We combine like terms...so anything with an x goes together and both constants go together.
5x equals negative 15. Divide 5 from both sides, and we get x equals negative 3.
Looks like our answer is B.
As in, "Bad managing decision."