Algebra · real student question

Solve 295 = x + 9.2 + (0.0152x)/(1 - 0.0152) for x.

Question

Solve for xx:

295=x+9.2+x0.015210.0152295 = x + 9.2 + \frac{x \cdot 0.0152}{1 - 0.0152}

Step-by-step solution

  1. Recognise that the fraction is linear in xx, not a new unknown. The term 0.0152x10.0152\dfrac{0.0152x}{1-0.0152} has xx only in the numerator, so it is simply xx times a constant. That means the whole equation is linear and can be solved by collecting like terms — no cross-multiplication is needed.

  2. Simplify the constant denominator.

    10.0152=0.98481-0.0152=0.9848

    so the fee term is 0.01520.9848x\dfrac{0.0152}{0.9848}\,x. Dividing:

    0.01520.9848=0.01543866...\frac{0.0152}{0.9848}=0.01543866...

    Keep the exact fraction 0.01520.9848=1529848=191231\tfrac{0.0152}{0.9848}=\tfrac{152}{9848}=\tfrac{19}{1231} to avoid rounding drift.

  3. Move the constant across and collect the xx terms. Subtract 9.29.2 from both sides:

    2959.2=x(1+191231)295-9.2=x\left(1+\frac{19}{1231}\right)

    285.8=12501231x285.8=\frac{1250}{1231}\,x

    since 1+191231=125012311+\tfrac{19}{1231}=\tfrac{1250}{1231}.

  4. Divide to isolate xx.

    x=285.812311250=351, ⁣819.81250=281.45584x=285.8\cdot\frac{1231}{1250}=\frac{351,\!819.8}{1250}=281.45584

    This is exact, not rounded — the fraction 12311250\tfrac{1231}{1250} terminates.

  5. Check by substituting back. With x=281.45584x=281.45584: the fee term is 0.0152281.45584/0.9848=4.27812.../0.9848=4.344160.0152\cdot 281.45584/0.9848=4.27812.../0.9848=4.34416, and 281.45584+9.2+4.34416=295.00000281.45584+9.2+4.34416=295.00000. The equation balances exactly, confirming x=281.45584x=281.45584.

Answer

x=281.45584x=281.45584

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