I have encountered a problem in algebra that involves doing this.
Problem
The sum of the squares of two numbers is 34, the first number being one less than twice the second number. Determine the numbers.
Answers: 3, 5
Let x - First Number
y - Second Number
Equations
$$ 1) \ x^2 + y^2 = 34 $$
$$ 2) \ x = 2y - 1 $$
I will show two different solutions.
Solution 1 - Quadratic in Terms of x
Manipulate Equation 2
$$ x + 1 = 2y $$
$$ \frac{x + 1}{2} = y $$
$$ y = \frac{x + 1}{2} $$
Substitute Equation 2 to Equation 1
$$ x^2 + (\frac{x + 1}{2})^2 = 34 $$ $$ x^2 + (\frac{(x + 1)^2}{4})^2 = 34 $$ $$ 4x^2 + (x + 1)^2 = 136 $$ $$ 4x^2 + x^2 + 2x + 1 = 136 $$ $$ 5x^2 + 2x - 135 = 0 $$
Solving for x
$$ x = 5, - \frac{27}{5} $$
$$ Disregard - \frac{27}{5}, use \ x = 5 $$ Substitute x = 5 to Equation 2 $$ y = \frac{5 + 1}{2} $$ $$ y = 3 $$
Solution 2 - Quadratic in Terms of y
Substitute Equation 2 to 1
$$ (2y - 1)^2 + y^2 = 36 $$ $$ 4y^2 - 4y + 1 + y^2 = 36 $$ $$ 5y^2 - 4y - 35 = 0 $$
Solving for y $$ y = 3, - \frac{11}{5} $$ $$ Disregard - \frac{11}{5}, use \ y = 3 $$ Substitute y = 3 to Equation 2 $$ x = 2(3) - 1 $$ $$ x = 5 $$
For both methods you obtain 3 and 5.
Through double checking manually, you can use the negative fraction root to find its pair value in terms of Equation 2. But substituting those values in Equation 1 will not produce the answer. This can feel very obvious through inspection since 34 is a whole number.
But what is the explanation on it not working out? For reference, I have already studied both Logarithmic Equations and Radical Equations which are both equations that produce incorrect solutions. For logarithms, it is because the answer does not comply with the domain that is set by the logarithms. For radicals, it has something to do with the signs of even roots being disregarded.
However this topic relies on a system of equations, particularly quadratic and linear which honestly I have not studied yet in detail. So I would also accept more references on these topics.
By the way, it is my first time using MathJax here. Apologies if the formatting looks bad
In fact, $(x,y)=\left(-\frac{11}{5},-\frac{27}{5}\right)$ satisfies both equations $x^2+y^2=34$ and $y=2x-1$. Perhaps you made an arithmetic error somewhere?
If you're in a context where "number" means "arbitrary real number", then the problem genuinely has two solutions. If there's some implicit restriction (e.g., the numbers have to be positive, integers, or both) you might be justified in discarding one of them.