Super-Hard SAT Math Question 97 | Free SAT Prep | Aniko

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Question 97·200 Super-Hard SAT Math Questions·Algebra

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(2k-1)(x-3) + (k+4)(2x+1) = (3k-2)(2x-1) - x - 7

In the given equation, $k$ is a positive integer. The equation has infinitely many solutions for $x$ . Which choice is the value of $k$ ?

When a parameter appears in a linear equation and you are told there are infinitely many solutions, don’t try to solve for $x$ . Instead, simplify both sides into the form $ax+b$ . For the equation to be true for every $x$ , the $x$ -coefficients must match and the constant terms must match; this creates simple equations in the parameter (here, $k$ ). Solve for $k$ and then select the matching choice.

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Expand carefully

Distribute on both sides, especially in products like $(2k-1)(x-3)$ and $(3k-2)(2x-1)$ .

What does “infinitely many solutions” mean here?

After simplifying, the left side and right side must be identical as expressions in $x$ (same $x$ -coefficient and same constant).

Match parts

Once each side is in the form $(\text{something})x + (\text{something})$ , set the $x$ -coefficients equal and also set the constants equal.

Desmos Guide

Graph each side as a line

Enter

y1=(2k-1)(x-3)+(k+4)(2x+1)
y2=(3k-2)(2x-1)-x-7

Desmos will create a slider for $k$ .

Restrict $k$ to positive integers

Click the slider settings and set the step to $1$ , and make sure you only test positive values of $k$ .

Look for complete overlap

Adjust $k$ until the two graphs lie exactly on top of each other (not just intersecting once). If they overlap, the equation is true for all $x$ , meaning infinitely many solutions.

Read off the matching $k$ value

The positive integer $k$ at which the two lines coincide is the value requested in the problem.

Step-by-step Explanation

Simplify the left side

Expand and combine like terms:

\begin{aligned} (2k-1)(x-3) &= (2k-1)x - 3(2k-1) \\ &= (2k-1)x + (-6k+3) \end{aligned}

\begin{aligned} (k+4)(2x+1) &= 2(k+4)x + (k+4) \\ &= (2k+8)x + (k+4) \end{aligned}

Add them:

\begin{aligned} \text{LHS} &= \big[(2k-1)+(2k+8)\big]x + \big[(-6k+3)+(k+4)\big] \\ &= (4k+7)x + (-5k+7) \end{aligned}

Simplify the right side

Expand and combine like terms:

\begin{aligned} (3k-2)(2x-1) &= 2(3k-2)x - (3k-2) \\ &= (6k-4)x + (-3k+2) \end{aligned}

Now include the remaining terms $-x-7$ :

\begin{aligned} \text{RHS} &= (6k-4)x - x + (-3k+2) - 7 \\ &= (6k-5)x + (-3k-5) \end{aligned}

Use the condition for infinitely many solutions

A linear equation has infinitely many solutions only if both sides are the same expression.

So match the coefficient of $x$ and match the constant term:

4k+7 = 6k-5

-5k+7 = -3k-5

Solve for $k$ and choose the answer

Solve either equation:

From $4k+7 = 6k-5$ :

12 = 2k \Rightarrow k = 6

(The constant-term equation gives the same result.)

Therefore, the correct choice is 6.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation