Question

4 from unit 4, lesson 6 solve each equation, and check your ans\\(2x + 4(3 - 2x) = \frac{3(2x + 2)}{6} + 4\\)

Explanation:

Step1: Simplify both sides

First, simplify the left - hand side (LHS) using the distributive property \(a(b + c)=ab+ac\). For \(2x + 4(3-2x)\), we have:
\(2x+4\times3-4\times2x=2x + 12-8x=- 6x + 12\)
Then, simplify the right - hand side (RHS). For \(\frac{3(2x + 2)}{6}+4\), we can first simplify \(\frac{3(2x + 2)}{6}\) by canceling out the common factor of 3 in the numerator and denominator: \(\frac{3(2x + 2)}{6}=\frac{2x + 2}{2}=x + 1\). So the RHS becomes \(x + 1+4=x + 5\)
Now our equation is \(-6x + 12=x + 5\)

Step2: Move the variable terms to one side

Subtract \(x\) from both sides of the equation:
\(-6x - x+12=x - x+5\)
\(-7x+12 = 5\)

Step3: Move the constant terms to the other side

Subtract 12 from both sides:
\(-7x+12 - 12=5 - 12\)
\(-7x=-7\)

Step4: Solve for x

Divide both sides by \(-7\):
\(\frac{-7x}{-7}=\frac{-7}{-7}\)
\(x = 1\)

Step5: Check the answer

Substitute \(x = 1\) into the original equation.
LHS: \(2\times1+4(3 - 2\times1)=2 + 4\times(3 - 2)=2+4\times1=2 + 4 = 6\)
RHS: \(\frac{3(2\times1+2)}{6}+4=\frac{3\times(2 + 2)}{6}+4=\frac{3\times4}{6}+4=\frac{12}{6}+4=2 + 4 = 6\)
Since LHS = RHS when \(x = 1\), the solution is correct.

Answer:

\(x = 1\)