Line is defined by . Line is defined by , where is a constant. For exactly one value of , the point of intersection of and lies on the line . Which of the following is that value of ?

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SAT Linear Equations in Two Variables Question 45 | Free SAT Prep | Aniko

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Question 45·Hard·Linear Equations in Two Variables

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Line $L_1$ is defined by $3x - 4y = 8$ . Line $L_2$ is defined by $(p-1)x + (p+2)y = 6$ , where $p$ is a constant.

For exactly one value of $p$ , the point of intersection of $L_1$ and $L_2$ lies on the line $y = 2x - 1$ .

Which of the following is that value of $p$ ?

When a parameter like $p$ appears in a line and the problem mentions a common intersection with other lines, first find any intersection point that does not involve the parameter—solve the system formed by the fixed lines. Then enforce that this concrete point lies on the parameterized line by substituting its coordinates into that line’s equation, which turns the problem into a simple one-variable linear equation in $p$ . This avoids messy multi-variable algebra and is both faster and less error-prone on the SAT.

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Use the lines without $p$ first

The point of intersection must lie on all three lines. Start by finding the intersection point of the two lines that have no $p$ : $3x - 4y = 8$ and $y = 2x - 1$ .

Solve the system $3x - 4y = 8$ and $y = 2x - 1$

Substitute $y = 2x - 1$ into $3x - 4y = 8$ to get an equation in $x$ only, then find $y$ from $y = 2x - 1$ .

Use the found point in the $L_2$ equation

Once you know the intersection point $(x,y)$ of $L_1$ and $y=2x-1$ , plug that $x$ and $y$ into $(p-1)x + (p+2)y = 6$ and solve the resulting linear equation for $p$ .

Check your algebra carefully

When you substitute and simplify, watch the negative signs and be sure to clear the fractions correctly (for example, multiply the entire equation by $5$ if denominators are $5$ ).

Desmos Guide

Graph the two fixed lines and find their intersection

Enter 3x - 4y = 8 and y = 2x - 1 into Desmos. Click on their point of intersection and note its coordinates (this point should be labeled automatically, such as A).

Add the parameterized line $L_2$ with a slider for $p$

In a new expression line, type (p-1)x + (p+2)y = 6. Desmos will prompt you to create a slider for p; accept it so that line $L_2$ appears and can move as you adjust $p$ .

Adjust $p$ until all three lines meet at one point

Drag the p slider until the graph of $(p-1)x + (p+2)y = 6$ passes exactly through the intersection point of 3x - 4y = 8 and y = 2x - 1. The value of p shown on the slider at that moment is the solution to the problem.

Step-by-step Explanation

Translate the geometric condition into equations

The phrase "the point of intersection of $L_1$ and $L_2$ lies on the line $y=2x-1$ " means there is one point $(x,y)$ that satisfies all three equations:

$3x - 4y = 8$ (line $L_1$ )
$(p-1)x + (p+2)y = 6$ (line $L_2$ )
$y = 2x - 1$ (third line)

To avoid dealing with $p$ too early, first find the intersection of the two lines that do not involve $p$ : $3x - 4y = 8$ and $y = 2x - 1$ .

Find the intersection of $L_1$ and $y=2x-1$

Use substitution with $y = 2x - 1$ in $3x - 4y = 8$ :

\begin{aligned} 3x - 4(2x - 1) &= 8 \\ 3x - 8x + 4 &= 8 \\ -5x + 4 &= 8 \\ -5x &= 4 \\ x &= -\frac{4}{5} \end{aligned}

Now find $y$ using $y = 2x - 1$ :

y = 2x - 1 = 2\left(-\frac{4}{5}\right) - 1 = -\frac{8}{5} - 1 = -\frac{13}{5}

So the common point of $L_1$ and $y=2x-1$ is $\left(-\frac{4}{5},\,-\frac{13}{5}\right)$ .

Force this point to lie on $L_2$ to get an equation in $p$

Now require that the same point lies on $L_2$ , whose equation is $(p-1)x + (p+2)y = 6$ .

Substitute $x = -\frac{4}{5}$ and $y = -\frac{13}{5}$ :

(p-1)\left(-\frac{4}{5}\right) + (p+2)\left(-\frac{13}{5}\right) = 6

Multiply both sides by $5$ to clear denominators:

\begin{aligned} -4(p-1) - 13(p+2) &= 30 \\ -4p + 4 - 13p - 26 &= 30 \\ -17p - 22 &= 30 \end{aligned}

Now you have a simple linear equation in $p$ .

Solve for $p$ and match the answer choice

Solve the equation from the previous step:

\begin{aligned} -17p - 22 &= 30 \\ -17p &= 52 \\ p &= -\frac{52}{17} \end{aligned}

Thus the value of $p$ that makes the intersection of $L_1$ and $L_2$ lie on $y = 2x - 1$ is $p = -\frac{52}{17}$ , which corresponds to choice C.

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Step-by-step Explanation

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Step-by-step Explanation