Question 658·Hard·Algebra

An x-intercept has $y=0$. The same point must satisfy both equations. Put $y=0$ in $3x+by=6$:

Divide by $3$:

So the lines meet at $(2,0)$.

Put the shared point $(2,0)$ in the first equation:

Combine terms:

Divide by $2$:

In standard form $Ax+By=C$, the slope is $-A/B$ when $B\ne0$. So the first line’s slope is $-a/4=-5/4$, and the second’s is $-3/b$. Here $b$ cannot be $0$: that would make the second line vertical, but the first line isn’t horizontal, so they couldn’t be perpendicular.

Perpendicular lines with these slopes have a product of $-1$:

Multiply the fractions:

Multiply by $4b$:

Type $15\sim-4b_{1}$ in Desmos. The subscript lets Desmos solve for $b_{1}$ rather than make a slider. Under PARAMETERS, it shows $b_{1}=-3.75$. Type $b_{1}$ on the next line and tap the fraction button to see $-\frac{15}{4}$.

Now $a=5$ and $b_{1}=b$. Type $4(5+b_{1})$ on the next Desmos line; it displays $5$. Keep the outside $4$: the question asks for $4(a+b)$, not $a+b$. Grid in 5.