Q. Solve: \(62x – 3 = 6 – 2x + 1\). Choices: \(x = -1, x = 0, x = 1, x = 4\).

Q: How do I simplify the right-hand side of 62x-3=6-2x+1?

Combine like terms: 6+1=7. The equation becomes 62x-3=7-2x.

Q: How do I solve 62x-3=7-2x for x?

Add 2x to both sides and add 3 to both sides: 64x = 10. Then x = \frac{10}{64} = \frac{5}{32}.

Q: Is the solution one of the choices x=-1,0,1,4?

No. The exact solution is x = \frac{5}{32} \approx 0.15625, which does not match any given choice.

Q: How can I check my solution quickly?

Substitute x = \frac{5}{32} into both sides. LHS: 62\left(\frac{5}{32}\right) - 3 = \frac{107}{16}. RHS: 7 - 2\left(\frac{5}{32}\right) = \frac{107}{16}. Equal, so the solution is correct.

Answer

see the next answer here to check \(62x-3=6-2x+1x\)

Combine like terms on the right: \(6-2x+1x=6-x\).

Solve: \(62x-3=6-x \Rightarrow 62x+x=6+3 \Rightarrow 63x=9 \Rightarrow x=\tfrac{1}{7}\).

Final result: \(x=\tfrac{1}{7}\) (none of the given choices \(-1,0,1,4\) is correct).

Detailed Explanation

Write down the equation and simplify the right-hand side by combining like terms (the constants 6 and 1):

\[62x – 3 = 6 – 2x + 1\]

\[62x – 3 = 7 – 2x\]
Eliminate the variable on the right by adding 2x to both sides to collect x-terms on the left:

\[62x + 2x – 3 = 7 – 2x + 2x\]

\[64x – 3 = 7\]
Isolate the term with x by adding 3 to both sides to remove the constant on the left:

\[64x – 3 + 3 = 7 + 3\]

\[64x = 10\]
Divide both sides by 64 to solve for x:

\[x = \frac{10}{64}\]

Simplify the fraction by dividing numerator and denominator by 2:

\[x = \frac{5}{32}\]
Optional check: substitute \(x = \frac{5}{32}\) into the original equation to verify equality.

Left-hand side:

\[62\left(\frac{5}{32}\right) – 3 = \frac{310}{32} – 3 = \frac{310}{32} – \frac{96}{32} = \frac{214}{32} = \frac{107}{16}\]

Right-hand side:

\[6 – 2\left(\frac{5}{32}\right) + 1 = 7 – \frac{10}{32} = 7 – \frac{5}{16} = \frac{112}{16} – \frac{5}{16} = \frac{107}{16}\]

Both sides equal \(\frac{107}{16}\), so the solution is correct.
Conclusion: The solution is

\[x = \frac{5}{32}\]

This value is not equal to any of the given choices x = -1, x = 0, x = 1, or x = 4.

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FAQs

How do I simplify the right-hand side of \(62x-3=6-2x+1\)?

Combine like terms: \(6+1=7\). The equation becomes \(62x-3=7-2x\).

How do I solve \(62x-3=7-2x\) for \(x\)?

Add \(2x\) to both sides and add 3 to both sides: \(64x = 10\). Then \(x = \frac{10}{64} = \frac{5}{32}\).

Is the solution one of the choices \(x=-1,0,1,4\)?

No. The exact solution is \(x = \frac{5}{32} \approx 0.15625\), which does not match any given choice.

How can I check my solution quickly?

Substitute \(x = \frac{5}{32}\) into both sides. LHS: \(62\left(\frac{5}{32}\right) - 3 = \frac{107}{16}\). RHS: \(7 - 2\left(\frac{5}{32}\right) = \frac{107}{16}\). Equal, so the solution is correct.

What common mistakes should I watch for solving this type of equation?

Typical errors: miscombining constants, sign mistakes when moving terms, forgetting to combine like terms on one side, and not simplifying fractions. Work step-by-step and re-check signs.

Could the original expression be interpreted differently, and would that change the answer?

How do I present the final answer clearly in multiple-choice situations?

Show the simplified equation, your algebra steps, and the final value \(x=\frac{5}{32}\). If that value isn't among choices, mark "none of the above" or note a misprint and explain your result.

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