Question

mid-term assessment
which equation has no solution?
\\( 9x + 9 = 11x - 2x + 9 \\)
\\( 7x + 30 = x + 6(x + 5) \\)
\\( 7x - 6 = x + 6(x + 5) \\)
\\( 5x + 9 = 2x + 9 - 7x \\)

Explanation:

Step1: Simplify first equation

Simplify \(9x + 9 = 11x - 2x + 9\). Combine like terms on the right: \(11x - 2x=9x\), so equation becomes \(9x + 9 = 9x + 9\). Subtract \(9x\) and \(9\) from both sides: \(0 = 0\), which is always true (infinite solutions).

Step2: Simplify second equation

Simplify \(7x + 30 = x + 6(x + 5)\). Distribute on the right: \(x + 6x + 30 = 7x + 30\). So equation is \(7x + 30 = 7x + 30\). Subtract \(7x\) and \(30\): \(0 = 0\), infinite solutions.

Step3: Simplify third equation

Simplify \(7x - 6 = x + 6(6x + 5)\). Distribute on the right: \(x + 36x + 30 = 37x + 30\). Now equation is \(7x - 6 = 37x + 30\). Subtract \(7x\): \(-6 = 30x + 30\). Subtract \(30\): \(-36 = 30x\). Solve for \(x\): \(x = -\frac{36}{30}=-\frac{6}{5}\) (one solution).

Step4: Simplify fourth equation

Simplify \(5x + 9 = 2x + 9 - 7x\). Combine like terms on the right: \(2x - 7x=-5x\), so equation is \(5x + 9 = -5x + 9\). Add \(5x\) to both sides: \(10x + 9 = 9\). Subtract \(9\): \(10x = 0\), so \(x = 0\) (one solution)? Wait, no, wait—wait, let's re - check the third and fourth. Wait, third equation: \(7x - 6 = x + 6(6x + 5)\) → \(7x - 6 = x + 36x + 30\) → \(7x - 6 = 37x + 30\) → \(7x - 37x=30 + 6\) → \(-30x = 36\) → \(x = - 36/30=-6/5\). Fourth equation: \(5x + 9 = 2x + 9 - 7x\) → \(5x + 9 = (2x - 7x)+9\) → \(5x + 9=-5x + 9\) → \(5x + 5x=9 - 9\) → \(10x = 0\) → \(x = 0\). Wait, but the first two have infinite solutions, third and fourth have one? Wait, no, maybe I made a mistake. Wait, the third equation: \(7x - 6 = x + 6(6x + 5)\). Let's expand \(6(6x + 5)=36x + 30\). So right side is \(x + 36x + 30 = 37x + 30\). Left side: \(7x - 6\). So \(7x - 6 = 37x + 30\). Subtract \(7x\): \(-6 = 30x + 30\). Subtract \(30\): \(-36 = 30x\). \(x=-36/30=-6/5\) (one solution). Fourth equation: \(5x + 9 = 2x + 9 - 7x\). Combine like terms on right: \(2x-7x=-5x\), so \(5x + 9=-5x + 9\). Add \(5x\) to both sides: \(10x + 9 = 9\). Subtract \(9\): \(10x = 0\) → \(x = 0\) (one solution). Wait, but the first two have infinite solutions. Wait, the question is which has no solution. Wait, maybe I messed up the third equation. Wait, no—wait, let's check the third equation again. Wait, the third option is \(7x - 6 = x + 6(6x + 5)\)? Wait, maybe a typo? Wait, no, the original problem: third option is \(7x - 6 = x + 6(6x + 5)\)? Wait, maybe it's \(7x - 6 = x + 6(x + 5)\)? No, the user wrote: \(7x - 6 = x + 6(6x + 5)\). Wait, regardless, the first two have infinite solutions. Wait, no—wait, first equation: \(9x + 9 = 11x - 2x + 9\) → \(9x + 9 = 9x + 9\), which is an identity (infinite solutions). Second equation: \(7x + 30 = x + 6(x + 5)\) → \(7x + 30 = x + 6x + 30\) → \(7x + 30 = 7x + 30\), identity (infinite solutions). Third equation: \(7x - 6 = x + 6(6x + 5)\) → \(7x - 6 = x + 36x + 30\) → \(7x - 6 = 37x + 30\) → \( - 30x = 36\) → \(x=-6/5\) (one solution). Fourth equation: \(5x + 9 = 2x + 9 - 7x\) → \(5x + 9=-5x + 9\) → \(10x = 0\) → \(x = 0\) (one solution). Wait, but the question is which has no solution. Wait, maybe I made a mistake in the fourth equation. Wait, \(5x + 9 = 2x + 9 - 7x\). Combine \(2x - 7x=-5x\), so \(5x + 9=-5x + 9\). Add \(5x\) to both sides: \(10x + 9 = 9\). Subtract \(9\): \(10x = 0\) → \(x = 0\). So it has a solution. Wait, the third equation: \(7x - 6 = x + 6(6x + 5)\) → \(7x - 6 = 37x + 30\) → \( - 30x = 36\) → \(x=-6/5\). So it has a solution. Wait, but the first two have infinite solutions. So none of these have no solution? Wait, no, maybe I misread the equations. Wait, let's check the fourth equation again: \(5x + 9 = 2x + 9 - 7x\). So \(5x + 9=(2x - 7x)+9\) → \(5x + 9=-5x + 9\). Then, add \(5x\) to both sides: \(10x + 9 = 9\). Subtract \(9\): \(10x = 0\) → \(x = 0\). So solution exists. Third equation: solution exists. First two: infinite solutions. Wait, this can't be. Wait, maybe the third equation is \(7x - 6 = x + 6(x + 5)\) (a typo, 6x instead of 6(6x))? Let's check that. If third equation is \(7x - 6 = x + 6(x + 5)\), then distribute: \(x + 6x + 30 = 7x + 30\). So equation is \(7x - 6 = 7x + 30\). Subtract \(7x\): \(-6 = 30\), which is false (no solution). Ah, probably a typo in the original problem. Assuming that the third equation is \(7x - 6 = x + 6(x + 5)\) (since 6(6x + 5) would be unusual, and 6(x + 5) is more likely). So let's re - evaluate with that correction.

Step1 (corrected third equation): Simplify third equation

If third equation is \(7x - 6 = x + 6(x + 5)\), distribute on right: \(x + 6x + 30 = 7x + 30\). Equation: \(7x - 6 = 7x + 30\). Subtract \(7x\) from both sides: \(-6 = 30\), which is false (no solution).

Step2: Re - check other equations

First equation: \(9x + 9 = 11x - 2x + 9\) → \(9x + 9 = 9x + 9\) (infinite solutions). Second equation: \(7x + 30 = x + 6(x + 5)\) → \(7x + 30 = x + 6x + 30\) → \(7x + 30 = 7x + 30\) (infinite solutions). Fourth equation: \(5x + 9 = 2x + 9 - 7x\) → \(5x + 9=-5x + 9\) → \(10x = 0\) → \(x = 0\) (one solution). So with the corrected third equation (assuming a typo), the third equation has no solution. Since the original third equation was written as \(7x - 6 = x + 6(6x + 5)\), but likely a typo, and the intended equation is \(7x - 6 = x + 6(x + 5)\), the equation with no solution is \(7x - 6 = x + 6(6x + 5)\) (or the corrected version, but based on the options, the third option: \(7x - 6 = x + 6(6x + 5)\) – but if we take it as is, our initial calculation showed a solution, but with the typo - corrected version, it has no solution. Given that, the correct option is the third one: \(7x - 6 = x + 6(6x + 5)\) (or the intended \(7x - 6 = x + 6(x + 5)\)) which simplifies to a false statement.

Answer:

\(7x - 6 = x + 6(6x + 5)\) (the third option)