When one letter is already alone, substitution beats elimination.
Solve the system
y = 2x − 3
4x + 3y = 21
y is already isolated, so substitute it: 4x + 3(2x − 3) = 21 gives 10x − 9 = 21 and x = 3. Then y = 2(3) − 3 = 3. Check: 12 + 9 = 21 ✓ C and D are the one-equation trap: (2, 1) and (4, 5) both sit on y = 2x − 3, but 4x + 3y gives 11 and 31, not 21. Always check in the equation you did not substitute into.
That is the whole reason to choose substitution here.
y = 2x − 3
y is by itself
4x + 3y = 21
both letters tangled
A letter already alone, or with a coefficient of 1 — substitute. Matching or opposite coefficients — eliminate. Either works on any system; one is just faster.
Every y in equation 2 becomes (2x − 3).
4x + 3y = 21
4x + 3(2x − 3) = 21
Writing 4x + 3 · 2x − 3 loses the multiplication on the −3. The bracket is what reminds you the 3 hits both terms.
One equation, one letter — which is the entire point.
4x + 3(2x − 3) = 21
from Step 2
4x + 6x − 9 = 21
distribute the 3
10x − 9 = 21
collect the x's
10x = 30
+ 9 both sides
x = 3
÷ 10 both sides
Use the simple equation, not the messy one.
y = 2x − 3
equation 1
y = 2(3) − 3
put x = 3 in
y = 3
6 − 3
Stopping at x = 3 is the most common mistake in this topic. A system asks for the point where the lines cross, and a point needs both numbers.
Two straight lines meet at exactly one point, so a system like this has exactly one answer.
If the letters all vanish and you are left with something false like 0 = 5, the lines are parallel and there is no solution. If you get 0 = 0, they are the same line and every point works.
Substitution still works — it just produces a quadratic. Where does the line y = x + 6 meet the parabola y = x²?
x² = x + 6
both equal y, so equal each other
x² − x − 6 = 0 → (x − 3)(x + 2) = 0
x = 3 or x = −2
y = 9 or y = 4
y = x + 6 for each
Exactly one meeting point means the quadratic has one repeated root: discriminant zero. For what k does y = 2x + k touch y = x² at exactly one point?
x² = 2x + k → x² − 2x − k = 0
a = 1, b = −2, c = −k
b² − 4ac = 4 + 4k = 0 → k = −1
one root: the line is tangent
A product works the same way. Solve x + y = 7 and xy = 12.
y = 7 − x → x(7 − x) = 12
substitute into the product
x² − 7x + 12 = 0 → x = 3 or 4
tidy and factor
(3, 4) and (4, 3)
pair each x with its own y
A quadratic gives two x values, and each has its own y. Mixing them — (3, 9) with (−2, 9) — gives points on only one of the graphs. Find y from the linear equation.
Solve 2x − y = 7 and 3x + 2y = 7.
−y = 7 − 2x → y = 2x − 7
isolate the letter with coefficient −1
3x + 2(2x − 7) = 7 → 7x − 14 = 7
substitute, brackets kept
x = 3, y = 2(3) − 7 = −1 → (3, −1)
check: 9 − 2 = 7 ✓
Different: nothing is alone at the start; pick the letter whose coefficient is 1 or −1 so no fractions appear.
From 2x − y = 7, writing y = 7 − 2x forgets to divide the whole line by −1. It gives x = −7 and y = 21, which fails 2x − y = 7: −14 − 21 = −35.
Solve y = 3x and 4x + 2y = 5.
4x + 2(3x) = 5 → 10x = 5
substitute
x = ½, y = 3(½) = 32
check: 2 + 3 = 5 ✓
Different: the crossing point does not have to land on whole numbers — fractions are a normal answer, not a sign of a slip.
Thirty tickets were sold, adults at ₱80 and children at ₱50, for ₱1,920. How many adult tickets?
a + c = 30 → c = 30 − a
count equation, isolated
80a + 50(30 − a) = 1920 → 30a + 1500 = 1920
money equation
a = 14, c = 16
1,120 + 800 = 1,920 ✓
Different: the system is not given — one equation counts things, the other counts their value.
| Also asked | The two equations |
|---|---|
| Digits: they add to 9, and reversing them makes the number 45 bigger | t + u = 9 and (10u + t) − (10t + u) = 45, so u − t = 5: t = 2, u = 7, the number is 27 |
| Ages: a mother is 4 times as old as her son; in 6 years she will be 3 times as old | m = 4s and m + 6 = 3(s + 6): 4s + 6 = 3s + 18, so s = 12 and m = 48 |
For what k do y = 3x − 1 and 6x − 2y = k have infinitely many solutions? Substitute and see what is left.
6x − 2(3x − 1) = k
brackets kept
6x − 6x + 2 = k → 2 = k
the x's vanish
k = 2: 2 = 2, true for every x
the same line: infinitely many
any other k: false, e.g. 2 = 5
parallel lines: no solution
Substitution items are often word problems (two numbers, coins, ages, digits) — name one unknown, write the other in terms of it, and answer the quantity asked. The harder ones pair a line with a parabola or a product, or ask for k. Always check the pair in the equation you did not substitute into.
The point has to satisfy both. Checking only the one you substituted into proves nothing.
Solve the system
x = 4y + 1
2x − 3y = 17
Substitute for x: 2(4y + 1) − 3y = 17 gives 5y + 2 = 17, so y = 3 and x = 4(3) + 1 = 13. C swaps the two values. Check: 26 − 9 = 17 ✓
Hint: this time it is x that sits alone, so x is what gets substituted. Keep the bracket when you replace it, and finish by finding both letters.
