Question

4411vacr v. algebra 1 sem a cr - a
linear equations and inequalities
what is the solution to the linear equation?
\\(\frac{2}{5} + p = \frac{4}{5} + \frac{3}{5}p\\)
\\(p = 2\\) \\(p = 10\\) \\(p = 8\\) \\(p = 1\\)

Explanation:

Step1: Subtract \(\frac{3}{5}p\) from both sides

To isolate the variable terms, we subtract \(\frac{3}{5}p\) from each side of the equation \(\frac{2}{5}+p=\frac{4}{5}+\frac{3}{5}p\). This gives us \(\frac{2}{5}+p - \frac{3}{5}p=\frac{4}{5}+\frac{3}{5}p-\frac{3}{5}p\), which simplifies to \(\frac{2}{5}+\frac{2}{5}p=\frac{4}{5}\).

Step2: Subtract \(\frac{2}{5}\) from both sides

Next, we subtract \(\frac{2}{5}\) from both sides to get \(\frac{2}{5}+\frac{2}{5}p-\frac{2}{5}=\frac{4}{5}-\frac{2}{5}\). Simplifying both sides, we have \(\frac{2}{5}p=\frac{2}{5}\).

Step3: Multiply both sides by \(\frac{5}{2}\)

To solve for \(p\), we multiply both sides of the equation \(\frac{2}{5}p=\frac{2}{5}\) by \(\frac{5}{2}\). So, \(p=\frac{2}{5}\times\frac{5}{2}\), which gives \(p = 1\)? Wait, no, wait. Wait, let's re - do step 1. Wait, original equation: \(\frac{2}{5}+p=\frac{4}{5}+\frac{3}{5}p\). Let's subtract \(\frac{3}{5}p\) from both sides: \(p-\frac{3}{5}p=\frac{4}{5}-\frac{2}{5}\). So \(\frac{2}{5}p=\frac{2}{5}\)? No, wait \(\frac{4}{5}-\frac{2}{5}=\frac{2}{5}\), and \(p-\frac{3}{5}p=\frac{2}{5}p\). Then \(\frac{2}{5}p=\frac{2}{5}\), so \(p = 1\)? But wait, let's check with \(p = 2\). Plug \(p = 2\) into left side: \(\frac{2}{5}+2=\frac{2 + 10}{5}=\frac{12}{5}\). Right side: \(\frac{4}{5}+\frac{3}{5}\times2=\frac{4 + 6}{5}=\frac{10}{5}=2=\frac{10}{5}\). Wait, \(\frac{12}{5}
eq\frac{10}{5}\). Wait, maybe I made a mistake. Let's start over.

Original equation: \(\frac{2}{5}+p=\frac{4}{5}+\frac{3}{5}p\)

Subtract \(\frac{3}{5}p\) from both sides: \(p-\frac{3}{5}p=\frac{4}{5}-\frac{2}{5}\)

\(p-\frac{3}{5}p=\frac{2}{5}p\), and \(\frac{4}{5}-\frac{2}{5}=\frac{2}{5}\). So \(\frac{2}{5}p=\frac{2}{5}\), so \(p = 1\)? But when \(p = 2\): left side \(\frac{2}{5}+2=\frac{2 + 10}{5}=\frac{12}{5}\), right side \(\frac{4}{5}+\frac{6}{5}=\frac{10}{5}=2=\frac{10}{5}\). Wait, that's not equal. Wait, maybe my first step was wrong. Let's try to get rid of fractions. Multiply both sides by 5: \(2 + 5p=4 + 3p\). Then subtract \(3p\) from both sides: \(2+2p = 4\). Then subtract 2: \(2p=2\), so \(p = 1\)? Wait, no, \(2 + 5p=4 + 3p\), subtract \(3p\): \(2 + 2p=4\), subtract 2: \(2p = 2\), so \(p = 1\). But when \(p = 1\), left side: \(\frac{2}{5}+1=\frac{7}{5}\), right side: \(\frac{4}{5}+\frac{3}{5}=\frac{7}{5}\). Oh, right! I made a mistake earlier when checking \(p = 2\). So the correct solution is \(p = 1\)? Wait, no, wait the options are \(p = 2\), \(p = 10\), \(p = 8\), \(p = 1\). Wait, let's do the equation again.

Original equation: \(\frac{2}{5}+p=\frac{4}{5}+\frac{3}{5}p\)

Multiply all terms by 5 to eliminate denominators: \(5\times\frac{2}{5}+5p=5\times\frac{4}{5}+5\times\frac{3}{5}p\)

Simplify: \(2 + 5p=4 + 3p\)

Subtract \(3p\) from both sides: \(2+5p - 3p=4+3p - 3p\)

Simplify: \(2 + 2p=4\)

Subtract 2 from both sides: \(2p=4 - 2\)

\(2p=2\)

Divide by 2: \(p = 1\). Wait, but let's check the options. The last option is \(p = 1\). So the correct answer is \(p = 1\).

Wait, but maybe I messed up the first step. Let's check with \(p = 1\):

Left side: \(\frac{2}{5}+1=\frac{2 + 5}{5}=\frac{7}{5}\)

Right side: \(\frac{4}{5}+\frac{3}{5}\times1=\frac{4 + 3}{5}=\frac{7}{5}\). They are equal. So the solution is \(p = 1\).

Answer:

\(p = 1\) (the option with \(p = 1\))