Question

3x=3
x=1
solve for bc
a---b---c, ab = x+1, bc=3x-1, ac=12
x+1 + 3x -1 =12
4x +1 -1 =12
4x -2 =12
+2 +2
4x=14
x=14/4
solve for x
a---b---c, ab=2(2x+1), bc=27, ac=12x-3
4x+2 +27 =12x-3
solve for x
a---b---c, ab=x-4, bc=2x-5, ac=5x-19
solve for ac
a---b---c, ab=2x, bc=2x+18, ac=6(x-1)

Explanation:

First problem (Solve for \( x \) in \( 3x = 3 \))

Step1: Divide both sides by 3

To solve for \( x \) in the equation \( 3x = 3 \), we divide both sides of the equation by 3. This is because we want to isolate \( x \), and dividing by the coefficient of \( x \) (which is 3) will give us the value of \( x \).
\[
\frac{3x}{3}=\frac{3}{3}
\]

Step2: Simplify both sides

Simplifying the left - hand side, \( \frac{3x}{3}=x \), and simplifying the right - hand side, \( \frac{3}{3} = 1 \).

Step1: Set up the equation for the length of \( AC \)

Since \( AC=AB + BC \), and \( AB=x + 1 \), \( BC = 3x-1 \), \( AC = 12 \), we substitute these values into the equation:
\[
(x + 1)+(3x-1)=12
\]

Step2: Combine like terms

Simplify the left - hand side of the equation. The \( + 1 \) and \( - 1 \) cancel out, and \( x+3x=4x \), so we get:
\[
4x=12
\]

Step3: Solve for \( x \)

Divide both sides of the equation \( 4x = 12 \) by 4:
\[
x=\frac{12}{4}=3
\]

Step4: Find the length of \( BC \)

Now that we know \( x = 3 \), we substitute \( x = 3 \) into the expression for \( BC \), which is \( 3x-1 \).
\[
BC=3\times3 - 1=9 - 1 = 8
\]

Step1: Set up the equation for the length of \( AC \)

Since \( AC=AB + BC \), and \( AB = 2(2x + 1)=4x + 2 \), \( BC = 27 \), \( AC=12x-3 \), we substitute these values into the equation:
\[
4x + 2+27=12x-3
\]

Step2: Combine like terms on the left - hand side

Simplify \( 4x+2 + 27 \) to get \( 4x+29 \). So the equation becomes:
\[
4x+29=12x-3
\]

Step3: Move the \( x \) terms to one side and constants to the other

Subtract \( 4x \) from both sides:
\[
29=8x-3
\]
Then add 3 to both sides:
\[
32 = 8x
\]

Step4: Solve for \( x \)

Divide both sides of the equation \( 32 = 8x \) by 8:
\[
x=\frac{32}{8}=4
\]

Answer:

\( x = 1 \)

Second problem (Solve for \( x \) and then \( BC \) where \( AB=x + 1 \), \( BC = 3x-1 \), and \( AC=12 \))