How many solutions?

Some SAT questions ask how many solutions there are, not what they are. In Desmos, you answer them by counting. An intersection is a point where the two graphs meet, and the whole idea fits in four words: one intersection, one solution.

Fast Desmos solution

Type each equation on its own line. The parabola and the line meet at $(1,0)$ and $(4,3)$. Two intersections mean two solutions, so the answer is C.

You didn't need those coordinates. The question asks how many, not where, so once you've found both intersections, you're done.

Here's how the picture turns into a count:

Have one equation instead of a system? Graph each side as its own expression. For $x^2=x+2$, graph $y=x^2$ and $y=x+2$. They meet twice, at $x=-1$ and $x=2$, and each of those $x$-values makes the equation true. Two intersections, two solutions.

A line that touches a parabola at exactly one point without crossing it is tangent to it.

Take the parabola

Its vertex, the lowest point, is at height $2$. Now picture three horizontal lines:

• $y=1$ runs below the parabola. They never meet, so there are zero solutions.
• $y=2$ touches the parabola at its vertex. One intersection, one solution.
• $y=5$ cuts through both sides. Two intersections, two solutions.

Here's a trap: the screen shows only part of each graph. A second intersection can hide outside the window. So before you trust a count, zoom out or drag the graph around, especially when a curve runs off the edge.

With two lines, there are only three possibilities:

• Different slopes: they cross once, so there's one solution.
• Same slope, different intercepts: they're parallel and never meet, so there are zero solutions.
• The same line written two ways: every point is shared, so there are infinitely many solutions.

The graph shows three equations:

• $y=2x+1$
• $y=2x-3$, parallel to the first
• $2y=4x+2$, which is the first equation with both sides doubled

The first two lines never meet. The first and third lie exactly on top of each other.

Standard form, $Ax+By=C$, hides the slopes, but you don't need them. Take $x+2y=4$ and $3x+6y=12$. Multiply the first equation by $3$ and you get the second one exactly. So they're the same line, with infinitely many solutions.

That's the test. Find the number that turns the first equation's $x$ and $y$ coefficients into the second's. Then check whether it works on the constant too. Keep $x+2y=4$ as the first equation, and two other things can happen:

• The number works on $x$ and $y$, but not on the constant. With $3x+6y=5$, the number is $3$, but $3\cdot4=12$, not $5$. The lines are parallel, so there are zero solutions.
• No number works. With $3x+5y=5$, multiplying by $3$ turns $x$ into $3x$, but it turns $2y$ into $6y$, not $5y$. The lines cross once, so there's one solution.

The number can be a fraction, too. Multiplying $6x+4y$ by $\frac{3}{2}$ gives $9x+6y$.

Overlapping graphs can also test whether two expressions are equivalent. That comes later, in Equivalent expressions by graph overlap.

The same idea works for one equation in $x$. If it has infinitely many solutions, its two sides are the same expression, true for every $x$. When a question asks for the constant that makes that happen, like the $k$ in $3(2x-k)+4=6x+k-8$, you're finding a constant in an identity, and find constants in equivalent expressions shows a quick Desmos fit for it.