Question 599·Hard·Algebra

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The graph shows two lines, $p$ and $q$ . The solution to the system of linear equations represented by the graph is the point $(x,y)$ where the two lines intersect.

Without solving for $x$ and $y$ , which choice gives the value of $k$ in the equation $15x + 2y = k$ for that solution $(x,y)$ ?

A system of linear equations can be shown as two lines even when their equations aren't printed. Use labeled points to write each line, then graph a proposed equation $15x+2y=k$ in Desmos. The right value makes that new line pass through the shared intersection. A line that meets only one of the original lines doesn't work.

Hints

Hint 1
A line's slope is its vertical change divided by its horizontal change. From the two labeled points on $p$ , how far does $y$ rise while $x$ moves right?
Hint 2
The point-slope equation $y-y_1=m(x-x_1)$ describes a line with slope $m$ through $(x_1,y_1)$ . Use a labeled point to graph each line in Desmos; you don't need the intersection's coordinates.
Hint 3
For any proposed value of $k$ , $15x+2y=k$ draws another line. The value works only if that line passes through the intersection of both original lines.

Step-by-step

Test the value at the shared intersection
Step 1
Find the slope of line $p$
From $A(-4,1)$ to $B(2,10)$ , the line rises $9$ units as it moves right $6$ units. Slope means rise divided by run, so $p$ has slope $\frac{9}{6}=\frac32$ .
Step 2
Graph line $p$
Use its slope and $A(-4,1)$ in point-slope form: $y-1=\frac32(x+4)$ . Type that equation in Desmos. It draws the line through both labeled points on $p$ .
Step 3
Find the slope of line $q$
From $C(-2,8)$ to $D(4,4)$ , $y$ falls $4$ while $x$ moves right $6$ . So $q$ has slope $\frac{-4}{6}=-\frac23$ .
Step 4
Graph line $q$
Use $C(-2,8)$ and that slope: $y-8=-\frac23(x+2)$ . Type it below the first equation. Desmos shows the shared crossing slightly left of the $y$ -axis; don't click for its coordinates.
Step 5
Rule out values that are too large
At the crossing, $x<0$ and $y<7$ . So $15x$ is negative and $2y<14$ , giving $k=15x+2y<14$ . Any proposed value above $14$ is too large.
Step 6
Test the smaller remaining value
Type $15x+2y=\frac{58}{13}$ . Its graph misses the shared crossing, so that value isn't $k$ . A proposed value works only when its line passes through the intersection of both original lines.
Step 7
Test the value that passes through
Type $15x+2y=\frac{146}{13}$ . Desmos shows this line going through the crossing of $p$ and $q$ . At their solution, $15x+2y$ is $\frac{146}{13}$ . Choice C.

Question 599·Hard·Algebra

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