Question 156·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Use substitution: the second equation says $y=2x-3$, so replace both occurrences of $y$ in the first equation:

Keep the parentheses around $2x-3$ when you square it.

Type the left side as $f(x)$ in Desmos. It graphs an upward-opening parabola, while $f(x)=m$ asks where that graph reaches the horizontal height $m$. A horizontal line meets this parabola exactly once only at its vertex, the lowest point. Any higher line meets both sides.

Click the vertex. Desmos labels it about $(0.4,-9.8)$, so the needed height $m$ is $-9.8$, not the $x$-coordinate $0.4$.

Type the choice $-49/5$ on another line; Desmos prints $-9.8$, matching the vertex’s height. So the value of $m$ that gives exactly one real solution is $-\frac{49}{5}$. Choice D.

Starting from the substituted equation, square the entire $2x-3$:

Distribute $6$ across both terms of $2x-3$:

Combine the $x^2$ terms, the $x$ terms, and the constants:

Each $x$ that satisfies this equation gives one $y$ through $y=2x-3$.

Subtract $m$ so the equation has the standard quadratic form $ax^2+bx+c=0$:

The discriminant is $b^2-4ac$, the part under the square root in the quadratic formula. Exactly one real root means it equals zero, because adding or subtracting zero gives the same root. Here $a=5$, $b=-4$, and $c=-9-m$, so:

Multiply through, keeping the negative sign on $-9-m$:

Add the constant terms:

Type $196+20m\sim0$ into Desmos. The $\sim$ tells Desmos to fit $m$; under PARAMETERS it shows $m=-9.8$.

Type $m$ on the next line and use its fraction button to see $-\frac{49}{5}$. So $m=-\frac{49}{5}$ gives the system exactly one real solution. Choice D.