Testinar Grade 6 math

Solving One-Step Equations

Students will build Grade 6 mastery of solving one-step equations using models, equations, and test-style reasoning.

CH03-07 Expressions and Equations

Directions Solve each problem. Show your reasoning on the page or on separate paper. Check the answer key only after finishing.

Solve for \(x\): \(x+7=15\)
1. \(x=7\)
2. \(x=8\)
3. \(x=15\)
4. \(x=22\)
Solve for \(x\): \(x+9=16\)
1. \(x=7\)
2. \(x=9\)
3. \(x=16\)
4. \(x=25\)
Solve for \(x\): \(x-5=12\)
1. \(x=7\)
2. \(x=12\)
3. \(x=17\)
4. \(x=19\)
Solve for \(x\): \(x-8=6\)
1. \(x=2\)
2. \(x=6\)
3. \(x=14\)
4. \(x=48\)
Solve for \(x\): \(3x=18\)
1. \(x=6\)
2. \(x=15\)
3. \(x=21\)
4. \(x=54\)
Solve for \(x\): \(5x=40\)
1. \(x=8\)
2. \(x=35\)
3. \(x=45\)
4. \(x=200\)
Solve for \(x\): \(\frac{x}{4}=6\)
1. \(x=1.5\)
2. \(x=10\)
3. \(x=24\)
4. \(x=2.4\)
Which equation is equivalent to \(x+6=13\)?
1. \(x=13-6\)
2. \(x=7\)
3. \(x=13\)
4. \(x=13+6\)

Teacher and parent reference

Answer key and explanations

B. \(x=8\)
Subtract \(7\) from both sides: \(x=15-7=8\).
A. \(x=7\)
Subtract \(9\) from both sides: \(x=16-9=7\).
C. \(x=17\)
Add \(5\) to both sides: \(x=12+5=17\).
C. \(x=14\)
Add \(8\) to both sides: \(x=6+8=14\).
A. \(x=6\)
Divide both sides by \(3\): \(x=18\div 3=6\).
A. \(x=8\)
Divide both sides by \(5\): \(x=40\div 5=8\).
C. \(x=24\)
Multiply both sides by \(4\): \(x=6\times 4=24\).
A. \(x=13-6\)
Subtract \(6\) from both sides to isolate \(x\). So \(x=13-6\).