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Algebra · real student question

Solve for X: (217*22 + (293 - 32X)*10) / (3.6*7*60) = (0.857X - 217/25.2)*28/60 + (0.857X - 103/25.2)*35/60 + (0.857X - 32X/25.2)*10/60.

Question

Solve for XX:

217⋅22+(293−32X)⋅103.6⋅7⋅60=(0.857X−21725.2)2860+(0.857X−10325.2)3560+(0.857X−32X25.2)1060.\frac{217\cdot 22+(293-32X)\cdot 10}{3.6\cdot 7\cdot 60}=\left(0.857X-\frac{217}{25.2}\right)\frac{28}{60}+\left(0.857X-\frac{103}{25.2}\right)\frac{35}{60}+\left(0.857X-\frac{32X}{25.2}\right)\frac{10}{60}.

Step-by-step solution

  1. Simplify the left-hand numerator. 217⋅22=4774217\cdot 22=4774 and (293−32X)⋅10=2930−320X(293-32X)\cdot 10=2930-320X, so the numerator is 7704−320X7704-320X. Everything to the left of the equals sign is linear in XX.

  2. Evaluate the left-hand denominator carefully. 3.6⋅7=25.23.6\cdot 7=25.2 and 25.2⋅60=151225.2\cdot 60=1512. This is the one place where an arithmetic slip changes the whole answer, so it is worth doing in two explicit multiplications. The left side is therefore 7704−320X1512=10721−40189X≈5.095238−0.211640X\frac{7704-320X}{1512}=\frac{107}{21}-\frac{40}{189}X\approx 5.095238-0.211640X.

  3. Expand the three right-hand terms. Using 25.2⋅60=151225.2\cdot 60=1512: the constant parts are −21725.2⋅2860=−60761512=−4.018519-\frac{217}{25.2}\cdot\frac{28}{60}=-\frac{6076}{1512}=-4.018519 and −10325.2⋅3560=−36051512=−2.384259-\frac{103}{25.2}\cdot\frac{35}{60}=-\frac{3605}{1512}=-2.384259, summing to −96811512=−46172≈−6.402778-\frac{9681}{1512}=-\frac{461}{72}\approx -6.402778. The third term has no constant part.

  4. Collect the coefficient of X on the right. The first two terms contribute 0.857(2860+3560)=0.857⋅1.05=0.899850.857\left(\frac{28}{60}+\frac{35}{60}\right)=0.857\cdot 1.05=0.89985, and the third contributes (0.857−3225.2)1060=(0.857−1.269841)⋅16=−0.068807\left(0.857-\frac{32}{25.2}\right)\frac{10}{60}=(0.857-1.269841)\cdot\frac16=-0.068807. Together the coefficient is 31413433780000≈0.831043\frac{3141343}{3780000}\approx 0.831043.

  5. Move all X terms to one side and constants to the other. From 5.095238−0.211640X=0.831043X−6.4027785.095238-0.211640X=0.831043X-6.402778 we get 5.095238+6.402778=(0.831043+0.211640)X5.095238+6.402778=(0.831043+0.211640)X, that is 11.498016=1.042683X11.498016=1.042683X.

  6. Divide to finish. X=11.4980161.042683=144875001313781≈11.0273X=\frac{11.498016}{1.042683}=\frac{14487500}{1313781}\approx 11.0273.

  7. Numerical check at X = 11.0273326. The left side is 7704−320(11.0273326)1512=4175.2531512=2.761411\frac{7704-320(11.0273326)}{1512}=\frac{4175.253}{1512}=2.761411, and the right side is 0.831043(11.0273326)−6.402778=2.7614100.831043(11.0273326)-6.402778=2.761410. Both sides agree to five decimal places.

Answer

X=144875001313781≈11.0273X=\frac{14487500}{1313781}\approx 11.0273

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