Question

1. in the equation below, what values will result in no solution?

$5x - 2(3x + 6) = \underline{quadquad} x - \underline{quadquad}$
\bigcirc a) $-5$ and $6$
\bigcirc b) $5$ and $12$
\bigcirc c) $-1$ and $6$
\bigcirc d) $-1$ and $12$

Explanation:

Step1: Simplify left side

First, simplify \(5x - 2(3x + 6)\). Distribute the -2: \(5x - 6x - 12\). Combine like terms: \(-x - 12\). So the left side is \(-x - 12\). The equation is \(-x - 12 = ax - b\) (where \(a\) is the coefficient of \(x\) and \(b\) is the constant on the right). For no solution, the coefficients of \(x\) must be equal (so the lines are parallel) and the constants must be different (so they never intersect).

Step2: Analyze each option

- Option A: \(a=-5\), \(b = 6\). Left side: \(-x - 12\), right side: \(-5x - 6\). Coefficients of \(x\) are different (\(-1 eq -5\)), so there is one solution. Eliminate A.
- Option B: \(a = 5\), \(b = 12\). Right side: \(5x - 12\). Coefficients of \(x\) (\(-1 eq5\)), one solution. Eliminate B.
- Option C: \(a=-1\), \(b = 6\). Right side: \(-x - 6\). Coefficients of \(x\) are equal (\(-1=-1\)), but constants: \(-12 eq -6\)? Wait, no: left side constant is -12, right side constant is -6? Wait, no, the equation is \(-x - 12 = -x - 6\). Subtract \(-x\) from both sides: \(-12 = -6\), which is false. Wait, but let's check Option D.
- Option D: \(a=-1\), \(b = 12\). Right side: \(-x - 12\). Then equation is \(-x - 12 = -x - 12\), which is an identity (infinite solutions). Wait, no, wait. Wait, original left side after simplifying is \(-x - 12\). So we need \(ax - b\) where \(a=-1\) (so same \(x\) coefficient) and \(b eq12\) for no solution? Wait, no, wait: if \(a=-1\) and \(b eq12\), then \(-x -12 = -x - b\) implies \(-12 = -b\) → \(b = 12\) for infinite, \(b eq12\) for no? Wait, no: let's re-express.

Wait, let's do it again. The left side is \(5x - 2(3x + 6) = 5x - 6x -12 = -x -12\). So the equation is \(-x -12 = mx - n\) (let \(m\) be the coefficient of \(x\), \(n\) the constant). For no solution, we need \(m = -1\) (same slope) and \(-n eq -12\) (different y-intercepts), so \(n eq12\)? Wait, no: the right side is \(mx - n\), so the constant term is \(-n\). Wait, the left side constant is -12, right side constant is -n. So for no solution, slopes equal (\(m = -1\)) and constants not equal (\(-12 eq -n\) → \(n eq12\)). Wait, but let's check the options:

Option C: \(m=-1\), \(n = 6\) (so right side is \(-x - 6\)). Then constants: -12 vs -6: not equal. So equation: \(-x -12 = -x -6\). Subtract \(-x\) from both sides: -12 = -6, which is false → no solution. Wait, but earlier I thought D was infinite. Wait, Option D: \(m=-1\), \(n=12\) → right side is \(-x -12\), so equation: \(-x -12 = -x -12\), which is always true (infinite solutions). So Option C: \(m=-1\), \(n=6\) → no solution. Wait, but let's recheck the simplification.

Wait, original left side: \(5x - 2(3x + 6) = 5x -6x -12 = -x -12\). So equation: \(-x -12 = ax - b\). So right side is \(ax - b\), so the constant term is \(-b\). So for no solution, \(a = -1\) (same coefficient of \(x\)) and \(-b eq -12\) → \(b eq12\). Let's check each option:

- A: \(a=-5\), \(b=6\): \(a eq -1\) → one solution.
- B: \(a=5\), \(b=12\): \(a eq -1\) → one solution.
- C: \(a=-1\), \(b=6\): \(a=-1\), \(b=6\) → \(-b = -6\). Left constant: -12, right constant: -6. Not equal → no solution.
- D: \(a=-1\), \(b=12\): \(-b = -12\). Left constant: -12, right constant: -12 → infinite solutions.

Ah, I made a mistake earlier. So Option C: \(a=-1\), \(b=6\) gives no solution. Wait, but let's confirm again.

Left side: \(-x -12\)

Right side for C: \(-1x - 6\) (since \(a=-1\), \(b=6\), so \(ax - b = -x -6\))

So equation: \(-x -12 = -x -6\)

Subtract \(-x\) from both sides: \(-12 = -6\), which is false. So no solution.

Option D: \(-x -12 = -x -12\), which is true for all x → infinite solutions.

So the correct answer is C? Wait, but let's check the options again. Wait, the problem says "what values will result in no solution". So let's re-express the left side correctly.

Wait, \(5x - 2(3x + 6) = 5x -6x -12 = -x -12\). So the equation is \(-x -12 = \text{(coefficient)}x - \text{(constant)}\). Let the coefficient be \(m\) and constant be \(n\) (so right side is \(mx - n\)). For no solution, we need \(m = -1\) (same slope) and \(-n eq -12\) (different y-intercepts), so \(n eq 12\).

Now check the options:

- A: \(m=-5\), \(n=6\) → \(m eq -1\) → one solution.
- B: \(m=5\), \(n=12\) → \(m eq -1\) → one solution.
- C: \(m=-1\), \(n=6\) → \(m=-1\), \(n=6\) (so \(-n = -6\)) → \(-12 eq -6\) → no solution.
- D: \(m=-1\), \(n=12\) → \(-n = -12\) → \(-12 = -12\) → infinite solutions.

So the correct option is C? Wait, but the original problem's options: Option C is -1 and 6, Option D is -1 and 12. Wait, maybe I messed up the sign of the constant. Wait, the right side is \(mx - n\), so the constant term is \(-n\). The left side constant is -12. So for the equation \(mx - n = -x -12\), we can rewrite it as \(mx - n + x +12 = 0\) → \((m + 1)x + (12 - n) = 0\). For no solution, coefficient of \(x\) is 0 (\(m + 1 = 0\) → \(m = -1\)) and constant term not 0 (\(12 - n eq 0\) → \(n eq 12\)). So \(m = -1\), \(n eq 12\). Looking at the options:

- A: \(m=-5\), \(n=6\) → \(m eq -1\) → one solution.
- B: \(m=5\), \(n=12\) → \(m eq -1\) → one solution.
- C: \(m=-1\), \(n=6\) → \(n eq12\) → no solution.
- D: \(m=-1\), \(n=12\) → \(n=12\) → infinite solutions.

Yes, so Option C is correct. Wait, but earlier I thought maybe D, but no. So the correct answer is C.

Answer:

C) \(-1\) and \(6\)