/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Solve the given systems of equat... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the given systems of equations algebraically. $$\begin{array}{l} y=2 x-1 \\ y=2 x^{2}+2 x-3 \end{array}$$

Short Answer

Expert verified
The solutions are \((x, y) = (1, 1)\) and \( (x, y) = (-1, -3) \).

Step by step solution

01

Equating the Expressions for y

Since both equations are equal to \( y \), set them equal to each other to find \( x \). So, solve for \( 2x - 1 = 2x^2 + 2x - 3 \).
02

Simplifying the Equation

Subtract \( 2x - 1 \) from both sides to eliminate \( x \) terms on one side: \[ 0 = 2x^2 + 2x - 3 - 2x + 1 \] This simplifies to \[ 0 = 2x^2 - 2 \].
03

Factor the Quadratic Equation

Factor the equation \( 2x^2 - 2 = 0 \). First, factor out the 2: \[ 2(x^2 - 1) = 0 \] Then, notice \( x^2 - 1 \) is a difference of squares: \[ 2(x - 1)(x + 1) = 0 \].
04

Solve for x

Set each factor in \( (x - 1)(x + 1) = 0 \) equal to zero to find the solutions for \( x \):\[ x - 1 = 0 \] gives \( x = 1 \)\[ x + 1 = 0 \] gives \( x = -1 \).
05

Solve for y using x Values

Substitute \( x = 1 \) and \( x = -1 \) back into the first equation \( y = 2x - 1 \) to find \( y \):For \( x = 1 \):\[ y = 2(1) - 1 = 1 \]For \( x = -1 \):\[ y = 2(-1) - 1 = -3 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
A quadratic equation is one of the most commonly encountered types of equations in algebra. It is a polynomial equation of the form ax^2 + bx + c = 0, where 'a', 'b', and 'c' are constants, and 'x' is the variable we aim to solve. The highest power of 'x' in a quadratic equation is 2, which is why these equations are called 'quadratic' (quad meaning square). In the context of systems of equations, as with the given exercise, one or both of the equations can especially be quadratic, indicating potential points where solutions for 'x' and 'y' can intersect. Quadratic equations have various methods of solving: one of the most common methods involves factoring, which often makes them simpler to solve.
Factoring
Factoring is an essential technique in algebra, particularly useful in simplifying expressions and solving quadratic equations. This process involves rewriting a quadratic equation into a product of its factors, making it easier to solve. In our example, after simplifying the expression as 2x^2 - 2 = 0, we factor by taking out the common factor of 2, which yields 2(x^2 - 1) = 0. The expression x^2 - 1 is then factored further as it represents the difference of squares. Factoring is not always straightforward, and sometimes a quadratic equation may not easily factor into integers. However, in many cases like this one, factoring helps identify the solutions by setting each factor to zero.
Difference of Squares
The term 'difference of squares' refers to a specific type of factoring where the expression is arranged as a^2 - b^2. This is a highly useful identity in algebra because it can always be expressed as (a - b)(a + b). In the example equation x^2 - 1, it is recognized as a difference of squares because it can be rewritten as x^2 - 1^2, which factors into (x - 1)(x + 1). Recognizing and applying the difference of squares formula can simplify many algebraic tasks, including solving equations, as it allows us to break down a seemingly complex quadratic expression into simpler linear factors.
Substitution Method
The substitution method is a practical approach for solving systems of equations. It involves solving one equation for one variable and substituting that expression into the other equation. This technique is evident in the exercise presented, where both equations are initially solved for 'y', allowing you to set the expressions equal to each other. By substituting one equation into the other, you reduce the system into a single equation with one variable, which simplifies the solving process. This method is very effective when dealing with linear and quadratic combinations like in our example, as it allows you to address each variable in turn, often paving the way for easier computation of values like 'x' and 'y'.

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