If x satisfies the linear equation \frac{3x - 1}{4} - \frac{2x + 3}{5} = \frac{x - 3}{4} +… — GRE General Test

x

satisfies the linear equation

\frac{3x - 1}{4} - \frac{2x + 3}{5} = \frac{x - 3}{4} + \frac{2}{5}

what is the value of

4x - 3

Solving the equation by grouping terms with like denominators gives

\frac{2x + 2}{4} = \frac{2x + 5}{5}

. Simplifying the left side to

\frac{x + 1}{2}

and cross-multiplying yields

5(x + 1) = 2(2x + 5)

, which simplifies to

5x + 5 = 4x + 10

, giving

x = 5

. Evaluating

4x - 3

x = 5

produces

4(5) - 3 = 17

, making this answer correct.

Group like fractional terms with denominator 4 on one side and denominator 5 on the other side of the equation.

\frac{3x - 1}{4} - \frac{x - 3}{4} = \frac{2x + 3}{5} + \frac{2}{5}

Grouping terms with common denominators simplifies algebraic combination.

Combine the numerators over their common denominators, carefully distributing signs.

\frac{(3x - 1) - (x - 3)}{4} = \frac{(2x + 3) + 2}{5} \implies \frac{2x + 2}{4} = \frac{2x + 5}{5}

Subtracting

(x - 3)

requires distributing the negative sign to yield

-x + 3

Simplify the left side fraction and cross-multiply to eliminate denominators.

\frac{x + 1}{2} = \frac{2x + 5}{5} \implies 5(x + 1) = 2(2x + 5)

Simplifying

\frac{2x+2}{4}

\frac{x+1}{2}

reduces computation before cross-multiplication.

Expand both sides and solve for

x

5x + 5 = 4x + 10 \implies 5x - 4x = 10 - 5 \implies x = 5

Subtracting

4x

and

5

from both sides isolates

x

Substitute

x = 5

into the required expression

4x - 3

4(5) - 3 = 20 - 3 = 17

The question asks for the value of

4x - 3

, not

x

alone.

Solving linear equations in one variable involving fractional terms and evaluating targeted algebraic expressions.

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