Problem 19 - Algebra 2 and Trigonometry

Question

In \(18-35,\) find each common solution algebraically. Express irrational roots in simplest radical form. $$ \begin{array}{l}{y=x^{2}+4 x} \\ {2 x-y=1}\end{array} $$

Step-by-Step Solution

Answer

The common solution is \((x, y) = (-1, -3)\).

Step 1: Substitute y in the Second Equation

The given equations are: 1. \( y = x^2 + 4x \) 2. \( 2x - y = 1 \)Substitute the expression for \(y\) from the first equation into the second equation:\[2x - (x^2 + 4x) = 1\]This simplifies to:\[-x^2 - 2x = 1\]which can be rewritten as:\[x^2 + 2x + 1 = 0\]by multiplying both sides by -1.

Step 2: Solve the Quadratic Equation

The equation we obtained is a quadratic equation:\[x^2 + 2x + 1 = 0\]This can be factored as:\[(x + 1)^2 = 0\]By solving this, we find:\[x + 1 = 0\]Therefore, we have:\[x = -1\]

Step 3: Find the Corresponding y

Using \(x = -1\), substitute back into the equation for \(y\):\[y = x^2 + 4x\]Substitute \(x = -1\):\[y = (-1)^2 + 4(-1)\]\[y = 1 - 4\]\[y = -3\]

Step 4: Verify the Solution

Check whether \(x = -1\) and \(y = -3\) satisfy both original equations:1. First equation: \(y = x^2 + 4x\)\[-3 = (-1)^2 + 4(-1)\]\[-3 = 1 - 4\]\[-3 = -3\] This is correct.2. Second equation: \(2x - y = 1\)\[2(-1) - (-3) = 1\]\[-2 + 3 = 1\]\[1 = 1\] This is correct. Both equations are satisfied.

Key Concepts

Substitution Method

The substitution method is a technique often used to find solutions for systems of equations. It's especially helpful when one equation is already solved for a single variable. In the given exercise, notice that the first equation, \( y = x^2 + 4x \), gives us \( y \) in terms of \( x \).

This allows us to substitute this expression for \( y \) into the second equation, \( 2x - y = 1 \). When you do this substitution, the equation becomes \( 2x - (x^2 + 4x) = 1 \). By replacing \( y \) we simplify to an equation with just one variable: \( x \).

Using substitution makes it more manageable by turning a system of equations into a single equation. This reduction is crucial for finding the solution.

Factoring Quadratic Equations

Factoring quadratic equations is a fundamental algebraic skill. It's about expressing a quadratic equation in the form \( ax^2 + bx + c = 0 \) as a product of its factors. In this solved exercise, we arrive at the quadratic equation \( x^2 + 2x + 1 = 0 \).

One common approach to factoring is identifying patterns or using techniques like the quadratic formula. Here, though, the equation is a perfect square trinomial, shown by:

• Notice: \( (x + 1)^2 = x^2 + 2x + 1 \).
• This means the equation can be rewritten as \( (x + 1)(x + 1) = 0 \).

Solving, we find \( x = -1 \). Once terms are in factored form, solving becomes simpler as you're finding values for which each factor equals zero. Mastery of factoring will enable you to tackle various algebraic problems with confidence.

Algebraic Verification

Once you find a potential solution, verifying it ensures it's correct. Algebraic verification involves substituting your results back into the original equations to see if they hold true.

In our example, we found \( x = -1 \) and a corresponding \( y = -3 \). We then check these values in both initial equations:

• First equation: \( y = x^2 + 4x \), substituting gives: \( -3 = (-1)^2 + 4(-1) \) which simplifies to \( -3 = -3 \).
• Second equation: \( 2x - y = 1 \), substituting leads: \( 2(-1) - (-3) = 1 \) simplifies to \( 1 = 1 \).

Both checks confirm the solution is accurate. Verification not only reassures correct calculations but also strengthens your understanding and confidence in solving such equations.