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Which value of [tex][tex]$p$[/tex][/tex] makes this equation true?

[tex]\[ \frac{p+1}{3} = \frac{p+1}{4} \][/tex]

A. [tex][tex]$p=-1$[/tex][/tex]
B. [tex][tex]$p=0$[/tex][/tex]
C. [tex][tex]$p=1$[/tex][/tex]
D. The equation has no solution.

Sagot :

To solve the equation [tex]\(\frac{p+1}{3} = \frac{p+1}{4}\)[/tex], we need to find the value of [tex]\(p\)[/tex] that satisfies this equality. Let's go through this step-by-step.

1. Write down the equation:

[tex]\[ \frac{p+1}{3} = \frac{p+1}{4} \][/tex]

2. Eliminate the fractions by finding a common denominator, which in this case is 12. Multiply every term by 12:

[tex]\[ 12 \cdot \frac{p+1}{3} = 12 \cdot \frac{p+1}{4} \][/tex]

3. Simplify both sides:

[tex]\[ 4(p+1) = 3(p+1) \][/tex]

4. Distribute the constants on both sides:

[tex]\[ 4p + 4 = 3p + 3 \][/tex]

5. Bring all terms involving [tex]\(p\)[/tex] to one side and constants to the opposite side. Subtract [tex]\(3p\)[/tex] and [tex]\(3\)[/tex] from both sides:

[tex]\[ 4p + 4 - 3p - 3 = 0 \][/tex]

[tex]\[ p + 1 = 0 \][/tex]

6. Solve for [tex]\(p\)[/tex]:

[tex]\[ p = -1 \][/tex]

7. Verify the solution: To ensure [tex]\(p = -1\)[/tex] is correct, substitute [tex]\(p = -1\)[/tex] back into the original equation:

[tex]\[ \frac{(-1)+1}{3} = \frac{(-1)+1}{4} \][/tex]

[tex]\[ \frac{0}{3} = \frac{0}{4} \][/tex]

[tex]\[ 0 = 0 \][/tex]

Since both sides of the equation are equal when [tex]\(p = -1\)[/tex], this value satisfies the equation.

Therefore, the value of [tex]\(p\)[/tex] that makes the equation true is:

[tex]\[ \boxed{-1} \][/tex]