With $y = 0$, the second equation becomes very simple. Use it to find $x$, then plug that $x$ into the first equation to solve for $a$.

Write both equations in the form $y = mx + c$ so you can see their slopes. How are the slopes of perpendicular lines related (what does their product equal)?

Once you know both $a$ and $b$, combine them to find $a + b$, then multiply by 4 as the question asks.

After solving the system algebraically, add two expressions in Desmos: a = [your value for a] and b = [your value for b] so Desmos treats them as constants.

Type the equations a*x + 4*y = 10 and 3*x + b*y = 6 into Desmos. Check that the graphs intersect at a point on the x-axis (where $y = 0$) and that the lines meet at a right angle, confirming they are perpendicular.

In a new line, type 4*(a + b). The numerical result that Desmos shows for this expression is the value you should use as your final answer.

If the two lines intersect at a point on the x-axis, that point has coordinates $(x, 0)$, so $y = 0$ at the intersection.

Substitute $y = 0$ into both equations:

• First equation: $ax + 4(0) = 10$, so $ax = 10$.
• Second equation: $3x + b(0) = 6$, so $3x = 6$.

From the second equation, solve for $x$:

Now plug $x = 2$ into $ax = 10$:

Rewrite each equation in slope-intercept form $y = mx + c$ to identify the slopes.

First equation: $ax + 4y = 10$

• Subtract $ax$ from both sides: $4y = -ax + 10$
• Divide by 4:

So the slope of the first line is $m_1 = -\dfrac{a}{4}$.

Second equation: $3x + by = 6$

• Subtract $3x$ from both sides: $by = -3x + 6$
• Divide by $b$:

So the slope of the second line is $m_2 = -\dfrac{3}{b}$.

Perpendicular lines have slopes that are negative reciprocals, so their product is $-1$:

Substitute $m_1 = -\dfrac{a}{4}$ and $m_2 = -\dfrac{3}{b}$:

Simplify the left side:

We already found $a = 5$, so substitute that in:

Solve for $b$:

Now use $a = 5$ and $b = -\dfrac{15}{4}$ to find $4(a + b)$.

First find $a + b$:

Now multiply by 4:

So the value of $4(a + b)$ is $5$.