Answered

Identify the first error Malik made when solving the equation:

[tex]\[
\frac{2}{5} x - 4 y = 10
\][/tex]

Malik's solution when [tex]\( x = 60 \)[/tex]:

[tex]\[
\begin{array}{l}
\frac{2}{5} x - 4 y = 10 \\
\frac{2}{5} \cdot 60 - 4(60) = 10 \\
\frac{2}{5} \cdot 60 - 240 = 10 \\
\frac{2}{5} x - 240 + 240 = 10 + 240 \\
\frac{5}{2}\left(\frac{2}{5} x\right) = \frac{5}{2}(250) \\
x = 265
\end{array}
\][/tex]

What error did Malik make first when solving the equation?

A. Malik did not multiply [tex]\(\frac{5}{2}(250)\)[/tex] correctly.
B. Malik added 240 to each side of the equation.
C. Malik did not multiply [tex]\(\frac{5}{2}\left(\frac{2}{5} x\right)\)[/tex] correctly.



Answer :

Let's solve the equation step-by-step carefully to identify the error Malik made.

Given the equation:
[tex]\[ \frac{2}{5} x - 4 y = 10 \][/tex]

We are also given that [tex]\( x = 60 \)[/tex], so substitute [tex]\( x = 60 \)[/tex] into the equation:

[tex]\[ \frac{2}{5} (60) - 4 y = 10 \][/tex]

Let's compute [tex]\(\frac{2}{5} (60)\)[/tex]:

[tex]\[ \frac{2}{5} \times 60 = \frac{120}{5} = 24 \][/tex]

Thus, the equation now becomes:

[tex]\[ 24 - 4 y = 10 \][/tex]

Next, isolate [tex]\( y \)[/tex] by subtracting 24 from both sides of the equation:

[tex]\[ -4 y = 10 - 24 \][/tex]

Simplify the right-hand side:

[tex]\[ -4 y = -14 \][/tex]

Now, solve for [tex]\( y \)[/tex] by dividing both sides by -4:

[tex]\[ y = \frac{-14}{-4} = 3.5 \][/tex]

Thus, the correct value of [tex]\( y \)[/tex] when [tex]\( x = 60 \)[/tex] is [tex]\( y = 3.5 \)[/tex].

Upon reviewing Malik's solution, he substituted [tex]\( y \)[/tex] with 60 in the second step, which was incorrect. The equation should have been solved for [tex]\( y \)[/tex] after substituting [tex]\( x \)[/tex], not after substituting [tex]\( y \)[/tex].

Therefore, the first error Malik made was in the second step:

[tex]\[ \frac{2}{5} x - 4(60) = 10 \][/tex]

This substitution of [tex]\( y = 60 \)[/tex] was incorrect.

Hence, the correct answer is:
Malik added 240 to each side of the equation.