Solve y = x², y = 2x
Worked out line by line the way a teacher would write it.
Answer
| Solutions | x = 0, y = 0; x = 2, y = 4 |
Step-by-step solution
10 steps-
1 The system, with the equations numbered\begin{aligned}y &= x^{2} & (1)\\ y &= 2 x & (2)\end{aligned}
-
2 Solve equation (1) for y
Substitution starts by solving one equation for one unknown, ideally one with coefficient 1 or −1. Then y can be replaced in the other equation.
y = x^{2} -
3 Substitute into equation (2)
Replacing the unknown by its expression leaves one equation with one unknown, which can be solved on its own.
x^{2} = 2 x -
4 Move every term to the left side so the right side is 0
Factoring and the quadratic formula both work on an equation of the form … = 0. Subtracting the right side from both sides gets there without changing the solutions.
x^{2} - 2 x = 0 -
5 This is a quadratic in standard form ax² + bx + c = 0
Every quadratic equation can be arranged as ax² + bx + c = 0. Reading off a = 1, b = -2 and c = 0, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -2,\quad c = 0 -
6 Every term contains x, so factor it out
Every term contains x, so x comes out as a common factor. Dividing both sides by x instead would be a mistake: it throws away the solution x = 0.
x \left(x - 2\right) = 0 -
7 Zero product property: a product is 0 only when one of its factors is 0
0 is the only number with this property: if a·b = 0, then a = 0 or b = 0. That is why the equation was first rearranged to … = 0 and factored: now each factor can be set to 0 on its own, giving a simpler equation for each.
x - 2 = 0\quad \text{or} \quad x = 0 -
8 Add 2 to both sides
An equation stays true as long as you do the same thing to both sides, like adding the same weight to both pans of a balance. Adding or subtracting the same amount on both sides moves a term to the other side, where it appears with the opposite sign.
x = 2Move the constant terms to the right.
-
9 Back-substitute x = 0
Once one unknown is known, putting its value into an earlier equation gives the other one.
y = 0^{2} = 0 -
10 Back-substitute x = 2y = 2^{2} = 4
Check by substitution
| Equation | Left side | Right side | |
|---|---|---|---|
| (1) | 0 | 0 | ✓ |
| (2) | 0 | 0 | ✓ |
| (1) | 4 | 4 | ✓ |
| (2) | 4 | 4 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.