1963 TMTA High School Algebra I Contest Problem 1

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Problem

If $4a - x = 3a - 4x,$ then $x =$

$\text{(A)} \quad -a \quad \text{(B)} \quad 7a/5 \quad \text{(C)} \quad a/5 \quad \text{(D)} \quad -a/3 \quad \text{(E) NOTA}$

Solution

Combine like terms: $3x = -a.$ We see that $x = -\frac{a}{3}$ and the answer is $\boxed{\text{(D)}}.$

See Also

1963 TMTA High School Mathematics Contests (Problems)
Preceded by
First Problem
TMTA High School Mathematics Contest Past Problems/Solutions Followed by
1963 TMTA High School Algebra I Contest Problem 2