Question

Solve the system: \(\left\\{\begin{array}{l}x-y=5 \\\ x-y^{2}=-1\end{array}\right.\)

Short answer

The solutions are \( (8, 3) \) and \( (3, -2) \).

Step-by-step solution

Identify the equations

The system of equations is: 1) \(x - y = 5\) 2) \(x - y^2 = -1\)

Express x in terms of y from the first equation

From the first equation, solve for \(x\): \(x = y + 5\)

Substitute x into the second equation

Replace \(x\) in the second equation with \(y + 5\): \( (y + 5) - y^2 = -1\)

Simplify the equation

Simplify \( (y + 5) - y^2 = -1\) to get: \( y + 5 - y^2 = -1 \). Rearrange to form a standard quadratic equation: \( -y^2 + y + 6 = 0 \), or equivalently \( y^2 - y - 6 = 0 \).

Solve the quadratic equation

Factorize the quadratic equation \( y^2 - y - 6 = 0 \): \( (y - 3)(y + 2) = 0 \). Thus, the solutions for \(y\) are \( y = 3 \) and \( y = -2 \).

Substitute y values back into the first equation to find x

For \( y = 3 \): \( x = y + 5 = 3 + 5 = 8 \). For \( y = -2 \): \( x = y + 5 = -2 + 5 = 3 \).

Write the solution pairs

Thus, the solution pairs are \( (x, y) = (8, 3) \) and \( (3, -2) \).

Key concepts

Linear Equations

Linear equations are the fundamental building blocks in algebra. They are called 'linear' because they graph as straight lines. These equations have variables raised only to the first power. For example, in the problem, the equation \(x - y = 5\) is a linear equation. The general form of a linear equation is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. Linear equations are straightforward to solve and form the basis for solving more complicated systems of equations.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2. This means the highest power of the variable is squared. For example, in our problem, the equation becomes a quadratic equation when simplified: \(y^2 - y - 6 = 0\). The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). Solving quadratic equations can be done through various methods like factoring, completing the square, or using the quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, we used factoring to find the values of \(y\).

Substitution Method

The substitution method involves solving one of the equations for a variable and then substituting that expression into the other equation. This method is particularly useful when one of the equations is already solved for one variable. In our problem, we first solved the linear equation \(x - y = 5\) for \(x\), giving us \(x = y + 5\). We then substituted \(x = y + 5\) into the quadratic equation \(x - y^2 = -1\), making it easier to solve.

Factoring

Factoring is a technique used to simplify complex polynomials. We break down a polynomial into a product of simpler polynomials. For instance, in our quadratic equation \(y^2 - y - 6 = 0\), we look for two numbers that multiply to \(-6\) and add to \(-1\). These numbers are \(-3\) and \(+2\), allowing us to factorize the equation as \((y - 3)(y + 2) = 0\). After factoring, we can set each factor equal to zero to find the values of \(y\).