Question

reasoning how is solving an equation with no solution similar to solving an equation that has an infinite number of solutions?14. critique reasoning your friend solved the equation $4x + 12x - 6 = 4(4x + 7)$ and got $x = 34$. what error did your friend make? what is the correct solution?$4x + 12x - 6 = 4(4x + 7)$$16x - 6 = 16x + 28$$16x - 16x - 6 = 16x - 16x + 28$$x - 6 = 28$$x - 6 + 6 = 28 + 6$$x = 34$16. classify the equation $6(x + 2) = 5(x + 7)$ as having no solution, or infinitely many solutions.17. solve $6x + 14x + 5 = 5(4x + 1)$. write a word problem that this equation, or any of its equivalent forms represents.

Explanation:

Step1: Analyze friend's error

When simplifying $16x - 16x - 6 = 16x - 16x + 28$, the friend incorrectly wrote $x - 6 = 28$ instead of $-6 = 28$. $16x-16x=0$, not $x$.

Step2: Correctly simplify the equation

Start with the original equation:
$4x + 12x - 6 = 4(4x + 7)$
Combine like terms and distribute:
$16x - 6 = 16x + 28$
Subtract $16x$ from both sides:
$16x - 16x - 6 = 16x - 16x + 28$
$-6 = 28$

Step3: Interpret the result

$-6 = 28$ is a false statement, so there is no value of $x$ that satisfies the equation.

Step1: Distribute to expand the equation

$6(x + 2) = 5(x + 7)$
$6x + 12 = 5x + 35$

Step2: Isolate the variable $x$

Subtract $5x$ from both sides:
$6x - 5x + 12 = 5x - 5x + 35$
$x + 12 = 35$
Subtract 12 from both sides:
$x + 12 - 12 = 35 - 12$
$x = 23$

Step3: Classify the solution type

The equation has exactly one unique solution, so it is neither "no solution" nor "infinitely many solutions".

Step1: Simplify both sides of the equation

Start with $6x + 14x + 5 = 5(4x + 1)$
Combine like terms and distribute:
$20x + 5 = 20x + 5$

Step2: Analyze the simplified equation

Subtract $20x$ from both sides:
$20x - 20x + 5 = 20x - 20x + 5$
$5 = 5$

Step3: Interpret the result

$5=5$ is a true statement, so all real numbers are solutions.

Step4: Create a word problem

Example: A café sells two combo meals. Combo A costs $6 per sandwich plus $14 per drink, plus a $5 service fee for any order. Combo B costs $5 for a set that includes a sandwich and a drink, where each set is priced as if it's $4 for the sandwich and $1 for the drink. How many meal sets (one sandwich + one drink) can you buy such that the total cost of Combo A equals Combo B?

Answer:

The friend incorrectly simplified $16x - 16x$ as $x$ instead of $0$. The equation has no solution.

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