Problem 68 - Algebra and Trigonometry

Question

Solve the system: \(\left\\{\begin{array}{l}{x-y=2} \\ {y^{2}=4 x+4}\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions are the points (8,6) and (0,-2).

Step 1: Isolate one variable

Isolate one of the variables from the linear equation, for instance, \(x\). By rearranging the first equation given \(x - y = 2\), the result is \(x = y + 2\)

Step 2: Substitute in the second equation

Take the expression of \(x\) and substitute it into the second equation in place of \(x\). The second equation is then transformed as follows: \(y^{2}=4(x+1) \Rightarrow y^{2}=4(y+2+1)\).

Step 3: Simplification and quadratic resolution

The substitution generates a quadratic equation. First simplify it and then solve it:\[y^{2}=4(y+3) \Rightarrow y^{2}-4y-12=0. \]Solving this quadratic equation by factoring gives \(y = 6\) or \(y = -2\).

Step 4: Solve for x

Now, substituting these found values of \(y\) back into the linear equation \(x = y + 2\), we can get the corresponding \(x\) values which are \(x = 8\) for \(y = 6\), and \(x = 0\) for \(y = -2\). Hence (8,6) and (0,-2) are solutions to the system.

Key Concepts

Substitution Method

When facing a system of equations, one efficient way to find a solution is to use the substitution method. This technique involves isolating one variable in one equation and then substituting the resulting expression into the other equation. It turns the system into a single-variable equation, allowing for straightforward solving.

For example, if you have a linear equation like \( x - y = 2 \) and a second equation involving both variables, isolate one variable. Let's pick \( x \) and rewrite the first equation as \( x = y + 2 \). This expression for \(x\) can now be substituted into any other equation that contains \(x\) to solve for \(y\). Once \(y\) is found, substitute it back into \( x = y + 2 \) to find the value of \(x\).

Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \). They are characterized by the presence of the highest exponent of 2 on the variable. These equations are fundamental in algebra and often appear in various areas of mathematics.

To solve a quadratic equation, one may employ several methods such as factoring, using the quadratic formula, completing the square, or even graphing. The solutions to these equations are the values of the variable that make the equation true and can be real or complex numbers.

Factoring Quadratics

Factoring is a crucial step when solving quadratic equations and involves rewriting the quadratic as a product of two binomial expressions. For example, a quadratic equation like \( y^2 - 4y - 12 = 0 \) can be factored into \( (y - 6)(y + 2) = 0 \).

When a quadratic is factored, it's based on the principle that if the product of two numbers is zero, then at least one of the numbers must be zero. Therefore, setting each factor equal to zero gives us the solutions to the equation. In the mentioned example, the solutions are \( y = 6 \) and \( y = -2 \), derived from the factors.

System of Linear and Quadratic Equations

A system that combines a linear equation with a quadratic equation presents a unique challenge. Solving such systems requires applying techniques for linear equations along with methods for solving quadratics.

In our example, the linear equation \( x - y = 2 \) is paired with a quadratic equation \( y^2 = 4x + 4 \). Once we isolate one variable in the linear equation and substitute it into the quadratic equation, we get a quadratic in one variable. By solving the quadratic, we get possible values for one variable, which are then substituted back into the linear equation to find the corresponding values of the second variable. The result is the set of ordered pairs that satisfy both equations in the system.