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Magazine Chapter 9: Problem 19
Solve the system of equations by applying the substitution method. $$\begin{array}{rr}x^{2}+x y-y^{2}= & 5 \\\x-y= & -1\end{array}$$
Short Answer
Expert verified
The solutions are \((x, y) = (3, 4)\) and \((x, y) = (-2, -1)\).
Step by step solution
01
Solve the Linear Equation for One Variable
We start with the linear equation given: \( x - y = -1 \). We solve this equation for \( x \) in terms of \( y \). Adding \( y \) to both sides gives \( x = y - 1 \). This expression for \( x \) will be used in the next step.
02
Substitute Expression into the Second Equation
Take the expression we found for \( x \) from Step 1, which is \( x = y - 1 \), and substitute it into the quadratic equation \( x^2 + xy - y^2 = 5 \). This gives us: \((y - 1)^2 + (y - 1)y - y^2 = 5\).
03
Expand and Simplify the Equation
Expand the terms: \( (y - 1)^2 = y^2 - 2y + 1 \) and \((y-1)y = y^2 - y \). Substitute these back into the equation: \( y^2 - 2y + 1 + y^2 - y - y^2 = 5 \). Simplify this to get \( y^2 - 3y + 1 = 5 \).
04
Solve the Resulting Quadratic Equation
Rearrange the terms: \( y^2 - 3y + 1 = 5 \) becomes \( y^2 - 3y - 4 = 0 \). This can be solved using the factorization method or quadratic formula. Factoring gives \((y - 4)(y + 1) = 0\), so the solutions for \( y \) are \( y = 4 \) and \( y = -1 \).
05
Calculate Corresponding Values of x
Using the solutions for \( y \), substitute back into \( x = y - 1 \). If \( y = 4 \), then \( x = 3 \). If \( y = -1 \), then \( x = -2 \). Thus, the solutions are \((x, y) = (3, 4)\) and \((x, y) = (-2, -1)\).
06
Verify Solutions in the Original Equations
Substitute \( (x, y) = (3, 4) \) into the original equations and verify: \( 3^2 + 3 \times 4 - 4^2 = 9 + 12 - 16 = 5 \) and \( 3 - 4 = -1 \). Similarly, test \( (x, y) = (-2, -1) \): \( (-2)^2 + (-2)(-1) - (-1)^2 = 4 + 2 - 1 = 5 \) and \( -2 - (-1) = -1 \). Both solutions satisfy the system.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Equations
A system of equations involves solving two or more equations simultaneously. The goal is to find a set of values for the variables that make all equations true. In this context, our goal is to solve for the variables \( x \) and \( y \) so that both the linear and quadratic equations are satisfied.
Here, the system of equations consists of a linear equation \( x - y = -1 \) and a quadratic equation \( x^2 + xy - y^2 = 5 \). By using the substitution method, we solve one equation for one variable and then substitute that expression into the other equation.
Here, the system of equations consists of a linear equation \( x - y = -1 \) and a quadratic equation \( x^2 + xy - y^2 = 5 \). By using the substitution method, we solve one equation for one variable and then substitute that expression into the other equation.
- This technique simplifies the system to a single equation in one variable, making it easier to solve.
- Once the main substitution step is done, the solution unfolds by solving the resulting equations.
- Finally, the found solutions are verified against the initial system to confirm their correctness.
Quadratic Equations
Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \). In our system of equations, the quadratic equation presents a crucial part. After substituting the linear expression into the quadratic equation, we arrive at \( y^2 - 3y - 4 = 0 \).
To solve quadratic equations, there are several methods that can be employed:
To solve quadratic equations, there are several methods that can be employed:
- Factorization: This involves expressing the quadratic as a product of two binomials. Here, we have \((y - 4)(y + 1) = 0\). The solutions, \(y = 4\) and \(y = -1\), are the values that make the equation zero.
- Quadratic Formula: If factorization isn't feasible, use the formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Understanding these methods allows students to tackle a broad range of problems beyond just this exercise.
Factorization
Factorization is a method of breaking down equations into simpler, solvable parts. In the system of equations exercise, we used factorization for our quadratic equation, \( y^2 - 3y - 4 = 0 \). This process involves finding two numbers that multiply to \(-4\) (the constant term) and add to \(-3\) (the linear coefficient).
The steps to factor are simple:
The steps to factor are simple:
- Identify the key values of the quadratic equation that need factoring.
- Use the relationships of these numbers to rewrite the equation in its product form, as in \((y - 4)(y + 1) = 0\).
- Solving this gives the roots \(y = 4\) and \(y = -1\). This is a straightforward way to solve quadratic equations, especially when they easily break into integer factors.
