Question

describe and correct any error a student may have made when solving the equation $0.15(y - 0.2)=2 - 0.5(1 - y)$.
the student first made an error when writing
(the expression on the
of that equation should be
the correct answer is $y = $
$0.15(y - 0.2)=2 - 0.5(1 - y)$
$0.15y - 0.03 = 2 - 0.5 + 0.5y$
$0.15y - 0.03 = 1.5 + 0.5y$
$1000(0.15y - 0.03)=1000(1.5 + 0.5y)$
$15y - 30 = 1500 + 500y$
$15y - 30 - 1500 - 1500 = 1500 + 500y - 1500 - 1500$
$-180 = 350y$
$y = \frac{-180}{350}$
(simplify your answers.)

Explanation:

Step1: Identify the error

Looking at the step where the student multiplies both sides by 1000, the right - hand side of the equation is \(2 - 0.5(1 - y)\). When multiplying by 1000, it should be \(1000\times2-1000\times0.5(1 - y)=2000 - 500(1 - y)\), but the student wrote \(100(1.5 + 0.5y)\) (which is incorrect). Also, when expanding \(0.15(y - 0.2)\), we get \(0.15y-0.03\), and when multiplying by 1000, it's \(150y - 30\), and the right - hand side when we correctly expand \(2-0.5(1 - y)=2-0.5 + 0.5y = 1.5+0.5y\), and multiplying by 1000 gives \(1500 + 500y\). But let's start from the original equation:

Original equation: \(0.15(y - 0.2)=2-0.5(1 - y)\)

Step2: Expand both sides

Left - hand side (LHS): Using the distributive property \(a(b - c)=ab - ac\), we have \(0.15y-0.15\times0.2=0.15y - 0.03\)

Right - hand side (RHS): Using the distributive property \(a - b(c - d)=a - bc+bd\), we have \(2-0.5\times1+0.5y=2 - 0.5+0.5y = 1.5 + 0.5y\)

So the equation becomes \(0.15y-0.03=1.5 + 0.5y\)

Step3: Multiply both sides by 1000 to eliminate decimals

Multiply each term by 1000:

\(1000\times(0.15y)-1000\times0.03=1000\times1.5+1000\times0.5y\)

\(150y-30 = 1500+500y\)

Step4: Move like terms to one side

Subtract \(150y\) from both sides:

\(150y-150y - 30=1500 + 500y-150y\)

\(- 30=1500 + 350y\)

Subtract 1500 from both sides:

\(-30 - 1500=350y\)

\(-1530 = 350y\) Wait, no, let's do it correctly. Wait, from \(0.15y-0.03=1.5 + 0.5y\)

Subtract \(0.15y\) from both sides:

\(-0.03=1.5 + 0.5y-0.15y\)

\(-0.03=1.5 + 0.35y\)

Subtract 1.5 from both sides:

\(-0.03-1.5 = 0.35y\)

\(-1.53=0.35y\)

Multiply both sides by 100 to eliminate decimals:

\(-153 = 35y\)

Then \(y=\frac{-153}{35}\approx - 4.37\)? Wait, maybe I made a mistake in the previous step. Wait, let's go back to the student's work. The student's error was in the multiplication factor. Let's use a better approach. Let's multiply both sides by 100 (to make it easier, since the decimals are to the hundredth place)

Original equation: \(0.15(y - 0.2)=2-0.5(1 - y)\)

Multiply both sides by 100:

\(15(y - 0.2)=200-50(1 - y)\)

Expand both sides:

\(15y-3 = 200-50 + 50y\)

\(15y-3=150 + 50y\)

Subtract \(15y\) from both sides:

\(-3=150 + 35y\)

Subtract 150 from both sides:

\(-3-150 = 35y\)

\(-153 = 35y\)

\(y=\frac{-153}{35}=-\frac{153}{35}\approx - 4.37\)

But looking at the student's work, when they multiplied by 1000, they had an error in the right - hand side. The correct expansion of the right - hand side after multiplying by 1000 (if we use 1000) should be \(1000\times(2-0.5(1 - y))=2000-500(1 - y)=2000 - 500+500y = 1500+500y\), and the left - hand side is \(1000\times(0.15y - 0.03)=150y-30\). Then:

\(150y-30=1500 + 500y\)

\(150y-500y=1500 + 30\)

\(-350y=1530\)

\(y=\frac{-1530}{350}=\frac{-153}{35}\) (dividing numerator and denominator by 10)

The student's error was in the step where they multiplied the right - hand side. They wrote \(100(1.5 + 0.5y)\) instead of \(1000(1.5 + 0.5y)\) (if they were multiplying by 1000) or \(100(1.5 + 0.5y)\) if they were multiplying by 100. Let's assume the student wanted to multiply by 100.

Correct step when multiplying by 100:

LHS: \(100\times(0.15y - 0.03)=15y-3\)

RHS: \(100\times(2-0.5(1 - y))=200-50(1 - y)=200 - 50+50y = 150+50y\)

So the equation is \(15y - 3=150+50y\)

Subtract \(15y\) from both sides: \(-3 = 150+35y\)

Subtract 150 from both sides: \(-153 = 35y\)

\(y=-\frac{153}{35}\)

The student's error was in the expansion of the right - hand side when multiplying by a factor (either wrong factor or wrong expansion). The correct answer for \(y\) is \(y =-\frac{153}{35}\) or approximately \(y\approx - 4.37\)

But let's re - examine the student's work:

Student's step: \(1000(0.15y - 0.03)=100(1.5 + 0.5y)\)

This is the error. They used different multipliers (1000 on the left and 100 on the right). It should be the same multiplier on both sides. If we use 1000 on the left, we should use 1000 on the right: \(1000(0.15y - 0.03)=1000(1.5 + 0.5y)\), which gives \(150y-30 = 1500+500y\)

Then, \(150y-500y=1500 + 30\)

\(-350y=1530\)

\(y=\frac{-1530}{350}=-\frac{153}{35}\)

Answer:

The student's error was using different multipliers (1000 on the left and 100 on the right) when eliminating decimals. The correct answer is \(y =-\frac{153}{35}\) (or \(y\approx - 4.37\))