Q. \(2(2x-4)=5(x-4)\).

Answer

Solve: \(2(2x-4)=5(x-4)\).

Step 1 – expand both sides:
\[
4x-8=5x-20.
\]

Step 2 – isolate \(x\): subtract \(4x\) from both sides,
\[
-8=x-20,
\]
then add 20:
\[
12=x.
\]

Final result: \(\boxed{x=12}\).

Detailed Explanation

Write the equation:\(2(2x-4)=5(x-4)\)
Apply the distributive property to both sides:\(2\cdot(2x)-2\cdot4 = 5\cdot x – 5\cdot4\)
So

\(4x-8 = 5x-20\)
Isolate the variable terms on one side:Subtract \(4x\) from both sides to move \(x\)-terms to the right:
\(4x-8-4x = 5x-20-4x\)

Which simplifies to

\(-8 = x-20\)
Solve for \(x\):Add \(20\) to both sides (inverse operation of subtraction):
\(-8+20 = x-20+20\)

\(12 = x\)

Thus \(x = 12\).
Check the solution by substitution:Left side: \(2(2x-4)=2(2\cdot12-4)=2(24-4)=2\cdot20=40\).
Right side: \(5(x-4)=5(12-4)=5\cdot8=40\).

Both sides equal \(40\), so the solution is correct.

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FAQs

What is the solution of /2(2x-4)=5(x-4)/?

Expand: /4x-8=5x-20/. Rearrange: /-8+20=5x-4x/ so /12=x/. Solution: /x=12/.

What are the steps to solve /2(2x-4)=5(x-4)/?

Distribute: /4x-8=5x-20/. Move variable terms one side: subtract /4x/: /-8=x-20/. -Add 20: /12=x/. Check by substitution.

How do I check the solution /x=12/?

Substitute: left /2(2(12)-4)=2(24-4)=2(20)=40/. Right /5(12-4)=5(8)=40/. Both equal 40, so /x=12/ is correct.

What common errors should I avoid?

Forgetting to distribute correctly, sign errors when moving terms, and arithmetic mistakes when combining constants. -Always expand parentheses and check signs when adding/subtracting.

Could this equation have no solution or infinitely many solutions?

Yes in general. If after simplifying you get a contradiction like /0=5/ there is no solution. If you get an identity like /0=0/ there are infinitely many. Here we got /x=12/, a unique solution.

Is there an alternative method besides distribution?

How does this look graphically?

Each side is a line: /y=4x-8/ and /y=5x-20/. The solution is their x-coordinate intersection. Solve /4x-8=5x-20/ to get intersection at /x=12/.

Why did I add 20 to both sides?

-After isolating variable terms you had /-8=x-20/. -Adding 20 cancels the constant on the right, isolating /x/ so /x=-8+20=12/.

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