Solve x² + y² = 13, x + y = 5
Worked out line by line the way a teacher would write it.
Answer
| Solutions | x = 3, y = 2; x = 2, y = 3 |
Step-by-step solution
14 steps-
1 The system, with the equations numbered\begin{aligned}x^{2} + y^{2} &= 13 & (1)\\ x + y &= 5 & (2)\end{aligned}
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2 Solve equation (2) for x
Substitution starts by solving one equation for one unknown, ideally one with coefficient 1 or −1. Then x can be replaced in the other equation.
x = 5 - y -
3 Substitute into equation (1)
Replacing the unknown by its expression leaves one equation with one unknown, which can be solved on its own.
y^{2} + \left(5 - y\right)^{2} = 13 -
4 Expand the brackets and collect like terms
Multiply out each bracket with the distributive property, a(b + c) = ab + ac, then combine like terms: terms with the same power of the variable, such as 3x and −x, add up to 2x.
2 y^{2} - 10 y + 25 = 13 -
5 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.
2 y^{2} - 10 y + 12 = 0 -
6 Divide both sides by the common factor 2
Every coefficient is divisible by 2. Dividing both sides by 2 gives smaller numbers and the same solutions, because 0 divided by 2 is still 0.
y^{2} - 5 y + 6 = 0 -
7 This is a quadratic in standard form ay² + by + c = 0
Every quadratic equation can be arranged as ay² + by + c = 0. Reading off a = 1, b = -5 and c = 6, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = -5,\quad c = 6 -
8 Factor the trinomial: find two numbers whose product is c = 6 and whose sum is b = -5
The goal is to write y² + by + c as (y + p)(y + q). Multiplying that out gives y² + (p + q)y + p·q, so p and q must multiply to c = 6 and add up to b = -5.
List the pairs of numbers whose product is 6, with their signs, and pick the pair whose sum is -5.
(-3) \cdot (-2) = 6,\qquad (-3) + (-2) = -5 -
9 Write the factored form\left(y - 3\right) \left(y - 2\right) = 0
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10 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.
y - 3 = 0\quad \text{or} \quad y - 2 = 0 -
11 Add 3 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.
y = 3Move the constant terms to the right.
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12 Add 2 to both sidesy = 2
Move the constant terms to the right.
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13 Back-substitute y = 2
Once one unknown is known, putting its value into an earlier equation gives the other one.
x = 5 - 2 = 3 -
14 Back-substitute y = 3x = 5 - 3 = 2
Check by substitution
| Equation | Left side | Right side | |
|---|---|---|---|
| (1) | 13 | 13 | ✓ |
| (2) | 5 | 5 | ✓ |
| (1) | 13 | 13 | ✓ |
| (2) | 5 | 5 | ✓ |
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