Where two functions meet, the algebra and the picture say the same thing.
A cell phone plan charges a flat monthly fee plus a per-minute rate. A competing plan charges differently. At some number of minutes, the two plans cost exactly the same. That crossover point is a solution to a system of equations — the input where both functions produce the same output. Finding that point is the whole problem.
Two functions f(x) and g(x) intersect wherever their outputs are equal. That condition is written as a single equation:
f(x)=g(x)
Every solution to this equation is an x-value where the graphs cross. The full solution is an ordered pair — you plug the x back into either function to get the y-coordinate.
Graphically, you find intersections by looking at the picture. Algebraically, you set the two expressions equal and solve. Both methods answer the same question. On the Regents, you need to know both.
Start with the simplest case: two lines. A linear-linear system has at most one solution — two distinct non-parallel lines cross exactly once.
Linear system: y = 2x + 1 and y = -x + 7
2x+1=−x+7
Set the two expressions equal. You want the x where both give the same y.
3x=6
Add x to both sides, subtract 1 from both sides.
x=2
Divide both sides by 3.
y=2(2)+1=5
Plug x = 2 into either equation to find y.
(2,5)✓
This is the point where both lines cross.
Interactive graph — scroll to zoom, drag to pan
The graph confirms it: both lines pass through (2,5).
Now raise the difficulty. When one equation is quadratic, the system can have zero, one, or two solutions. A line can miss a parabola entirely, graze it at exactly one point, or cut through it at two points.
To solve a linear-quadratic system, substitute. If one equation gives you y as a linear expression in x, replace y in the quadratic equation with that expression. Then solve the resulting quadratic.
Linear-quadratic system: y = x + 2 and y = x² - 4
x+2=x2−4
Both expressions equal y, so set them equal to each other.
0=x2−x−6
Move everything to one side. Subtract x and subtract 2 from both sides.
0=(x−3)(x+2)
Factor the quadratic. You need two numbers that multiply to -6 and add to -1.
x=3orx=−2
Set each factor equal to zero and solve.
y=(3)+2=5andy=(−2)+2=0
Substitute both x-values into the linear equation — it's simpler than the quadratic.
(3,5) and (−2,0)✓
Two intersection points. Verify each in both original equations if the problem asks.
Interactive graph — scroll to zoom, drag to pan
The graph shows two crossing points. One is at x=3 in the positive region, the other at x=−2 on the negative side. The algebra found both.
What if the line doesn't cross the parabola at all? The discriminant of the resulting quadratic will be negative — no real solutions. If the discriminant is zero, the line is tangent to the parabola: exactly one intersection.
Algebra II also brings circles into systems. A circle centered at the origin with radius r has equation:
x2+y2=r2
A circle is not a function, but it still appears in systems. Pair it with a line, and you have a linear-conic system. The approach is the same: solve one equation for one variable, substitute into the other.
Circle and line: x² + y² = 25 and y = x + 1
x2+(x+1)2=25
Substitute y = x + 1 into the circle equation. Replace every y with the linear expression.
x2+x2+2x+1=25
Expand (x+1)² using FOIL or the identity (a+b)² = a² + 2ab + b².
2x2+2x−24=0
Combine like terms, then subtract 25 from both sides.
x2+x−12=0
Divide every term by 2 to simplify before factoring.
(x+4)(x−3)=0
Factor. You need two numbers that multiply to -12 and add to 1.
x=−4orx=3
Zero-product property.
y=(−4)+1=−3andy=(3)+1=4
Substitute both x-values into y = x + 1.
(−4,−3) and (3,4)✓
Both points lie on the circle and on the line. You can check: (-4)² + (-3)² = 16 + 9 = 25.
Interactive graph — scroll to zoom, drag to pan
Two points on a circle with radius 5 — you can verify each satisfies the circle equation.
When the Regents asks you to solve a system graphically, sketch both curves carefully, label the intersection points, and write the coordinates. When it asks algebraically, show all substitution steps and all factoring. A system with two solutions needs both — leaving one out costs points.
Practice Questions
y=3x−1 and y=x+5
y=x2−3x and y=2x−6
x2+y2=100 and y=−2x+10
Regents Corner
On Part II and Part III of the Algebra II Regents, system problems are common and partial credit is real. Showing the substitution step explicitly — even if it seems obvious — protects your score. Showing only the final answers with no work earns no credit even if the answers are correct.
After factoring and finding two x-values, students plug both back into the quadratic equation instead of the linear one. The linear equation is simpler and less likely to produce arithmetic errors. Either equation gives the correct y, but using the harder one increases your chance of a sign mistake that costs you the point.