Answered

Which shows the correct solution of the equation [tex]\frac{1}{2} a+\frac{2}{3} b=50[/tex], when [tex]b=30[/tex]?

[tex]\[
\begin{array}{l}
\frac{1}{2} a+\frac{2}{3}(30)=50 \\
\frac{1}{2} a+20=50 \\
\frac{1}{2} a+20-20=50-20 \\
\frac{1}{2} a=30 \\
2\left(\frac{1}{2} a\right)=2(30) \\
a=60
\end{array}
\][/tex]

[tex]\[
\begin{aligned}
\frac{1}{2} a+\frac{2}{3}(30) & =50 \\
\frac{1}{2} a+20 & =50
\end{aligned}
\][/tex]

[tex]\[
\begin{aligned}
\frac{1}{2} a+20-20 & =50-20 \\
\frac{1}{2} a & =30 \\
a & =60
\end{aligned}
\][/tex]



Answer :

GinnyAnswer
To solve the equation [tex]\(\frac{1}{2}a + \frac{2}{3}b = 50\)[/tex] when [tex]\(b = 30\)[/tex], follow these steps:

1. Substitute [tex]\(b = 30\)[/tex] into the equation:
[tex]\[ \frac{1}{2}a + \frac{2}{3}(30) = 50 \][/tex]

2. Calculate [tex]\(\frac{2}{3} \times 30\)[/tex]:
[tex]\[ \frac{2}{3} \times 30 = 20 \][/tex]

3. Substitute the value back into the equation:
[tex]\[ \frac{1}{2}a + 20 = 50 \][/tex]

4. To isolate [tex]\(\frac{1}{2}a\)[/tex], subtract 20 from both sides of the equation:
[tex]\[ \frac{1}{2}a = 50 - 20 \][/tex]
[tex]\[ \frac{1}{2}a = 30 \][/tex]

5. Solve for [tex]\(a\)[/tex] by multiplying both sides of the equation by 2:
[tex]\[ a = 30 \times 2 \][/tex]
[tex]\[ a = 60 \][/tex]

Thus, the correct solution shows that when [tex]\(b = 30\)[/tex], the value of [tex]\(a\)[/tex] is [tex]\(60\)[/tex].

Other Questions