Expressions and Equations: Solving Linear Equations in One Variable (CCSS.8.EE.7)
Common Core 8th Grade Math · Learn by Concept
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Common Core 8th Grade Math › Expressions and Equations: Solving Linear Equations in One Variable (CCSS.8.EE.7)
What is the solution to $3(x-2)+4=19$?
$\tfrac{17}{3}$
$21$
$7$
$\tfrac{29}{3}$
Explanation
Distribute: $3(x-2)+4=19 \Rightarrow 3x-6+4=19$. Combine like terms: $3x-2=19$. Add 2 to both sides: $3x=21$. Divide by 3: $x=7$.
Solve the equation: $\dfrac{2}{3}(x+6)-\dfrac{1}{2}(x-4)=5$
$-6$
$18$
$-1$
$6$
Explanation
Distribute: $\tfrac{2}{3}x+4-\left(\tfrac{1}{2}x-2\right)=5$. Combine like terms: $\left(\tfrac{2}{3}-\tfrac{1}{2}\right)x+6=5 \Rightarrow \tfrac{1}{6}x+6=5$. Subtract 6: $\tfrac{1}{6}x=-1$. Multiply by 6: $x=-6$.
Solve: $5-2(3x+1)=-(x-7)+8$
$\tfrac{2}{5}$
$-\tfrac{9}{5}$
$-12$
$-\tfrac{12}{5}$
Explanation
Distribute: LHS $=5-6x-2=3-6x$. RHS $=-(x-7)+8=-x+7+8=-x+15$. Set equal: $3-6x=-x+15$. Add $6x$: $3=5x+15$. Subtract $15$: $-12=5x$. Divide by 5: $x=-\tfrac{12}{5}$.
Determine the solution set of $4(x-3)+2x=6x-12$.
$x=2$
Infinitely many solutions
No solution
$x=-2$
Explanation
Distribute and simplify: LHS $=4x-12+2x=6x-12$. Thus $6x-12=6x-12$, which is always true. This is an identity ($a=a$), so there are infinitely many solutions.
What is the solution to $\frac{1}{2}(4x - 6) - \frac{3}{4}(x + 2) = \frac{5}{2}$?
$\frac{14}{5}$
$\frac{28}{5}$
$7$
$-\frac{28}{5}$
Explanation
Distribute: $\tfrac{1}{2}(4x - 6)=2x-3$ and $\tfrac{3}{4}(x+2)=\tfrac{3}{4}x+\tfrac{3}{2}$. So $2x-3-\big(\tfrac{3}{4}x+\tfrac{3}{2}\big)=\tfrac{5}{2}$. Combine like terms: $(2-\tfrac{3}{4})x+(-3-\tfrac{3}{2})=\tfrac{5}{2}\Rightarrow \tfrac{5}{4}x-\tfrac{9}{2}=\tfrac{5}{2}$. Add $\tfrac{9}{2}$: $\tfrac{5}{4}x=\tfrac{14}{2}=7$. Multiply by $\tfrac{4}{5}$: $x=\tfrac{28}{5}$.
What is the solution to $3(x - 2) + 4 = 19$?
$\frac{17}{3}$
$-7$
$21$
$7$
Explanation
Distribute: $3x-6+4=19$. Combine like terms: $3x-2=19$. Add $2$: $3x=21$. Divide by $3$: $x=7$.
Solve the equation: $3(x + 2) - (x - 1) = 2x + 7$.
Infinitely many solutions
$0$
No solution
$1$
Explanation
Distribute: $3x+6 - x + 1 = 2x + 7$. Combine like terms: $2x+7=2x+7$. Subtract $2x$: $7=7$. The identity $a=a$ is always true, so there are infinitely many solutions.
What is the solution to $2(3x - 5) - 4(x - 2) = 2x - 6$?
$2$
Infinitely many solutions
No solution
$-2$
Explanation
Distribute: $6x-10 - 4x + 8 = 2x - 6$. Combine like terms: $2x - 2 = 2x - 6$. Subtract $2x$: $-2 = -6$. The false statement $a=b$ (with $a\neq b$) means there is no solution.
What is the solution to $3(x-2)+4=19$?
$x=17$
$x=5$
$x=7$
$x=-7$
Explanation
Distribute and simplify: $3(x-2)+4=19 \Rightarrow 3x-6+4=19 \Rightarrow 3x-2=19$. Add 2 to both sides: $3x=21$. Divide by 3: $x=7$. Common errors: forgetting to divide by 3 gives $x=21$ or $x=17$; distribution/sign mistakes can lead to $x=5$ or $x=-7$.
Solve $\dfrac{3}{4}(2x-6)-\dfrac{5}{2}=\dfrac{1}{2}(x+2)$.
$x=8$
$x=4$
$x=-8$
$x=16$
Explanation
Distribute: $\tfrac{3}{4}(2x-6)=\tfrac{3}{2}x-\tfrac{9}{2}$ and $\tfrac{1}{2}(x+2)=\tfrac{1}{2}x+1$. So $\tfrac{3}{2}x-\tfrac{9}{2}-\tfrac{5}{2}=\tfrac{1}{2}x+1 \Rightarrow \tfrac{3}{2}x-7=\tfrac{1}{2}x+1$. Subtract $\tfrac{1}{2}x$: $x-7=1$. Add 7: $x=8$. Distractors reflect combining fractions incorrectly ($x=4$), sign mistakes ($x=-8$), or arithmetic/scale errors ($x=16$).