Solve x² − y = 1, x + y = 5
Worked out line by line the way a teacher would write it.
Answer
| Solutions | x = −3, y = 8; x = 2, y = 3 |
Step-by-step solution
12 steps-
1 The system, with the equations numbered\begin{aligned}x^{2} - y &= 1 & (1)\\ x + y &= 5 & (2)\end{aligned}
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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} - 1 -
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 + x^{2} - 1 = 5 -
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} + x - 6 = 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 = 1 and c = -6, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = 1,\quad c = -6 -
6 Factor the trinomial: find two numbers whose product is c = -6 and whose sum is b = 1
The goal is to write x² + bx + c as (x + p)(x + q). Multiplying that out gives x² + (p + q)x + p·q, so p and q must multiply to c = -6 and add up to b = 1.
List the pairs of numbers whose product is -6, with their signs, and pick the pair whose sum is 1.
(-2) \cdot 3 = -6,\qquad (-2) + 3 = 1 -
7 Write the factored form\left(x - 2\right) \left(x + 3\right) = 0
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8 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 + 3 = 0 -
9 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.
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10 Subtract 3 from both sidesx = -3
Move the constant terms to the right.
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11 Back-substitute x = -3
Once one unknown is known, putting its value into an earlier equation gives the other one.
y = \left(-3\right)^{2} - 1 = 8 -
12 Back-substitute x = 2y = 2^{2} - 1 = 3
Check by substitution
| Equation | Left side | Right side | |
|---|---|---|---|
| (1) | 1 | 1 | ✓ |
| (2) | 5 | 5 | ✓ |
| (1) | 1 | 1 | ✓ |
| (2) | 5 | 5 | ✓ |
Change a number or try your own problem: the solver works it out the same way, with a graph and hint mode.