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Chapter 9: Problem 13

Solve the system of equations by applying the substitution method. $$\begin{aligned}&x+y=2\\\&x^{2}+y^{2}=2\end{aligned}$$

Short Answer

Expert verified
The solution is \( (x, y) = (1, 1) \). Both equations are satisfied.

Step by step solution

01

Solve for One Variable

Start with the first equation: \( x + y = 2 \). Solve for \( y \), yielding \( y = 2 - x \). This expression allows us to substitute \( y \) in terms of \( x \) into the second equation.
02

Substitute and Simplify

Substitute \( y = 2 - x \) into the second equation \( x^2 + y^2 = 2 \). So, we have: \[ x^2 + (2 - x)^2 = 2. \] Expanding \((2-x)^2\) gives \(4 - 4x + x^2\). The equation becomes \( x^2 + 4 - 4x + x^2 = 2 \). Simplify this to \( 2x^2 - 4x + 4 = 2 \).
03

Set the Equation to Zero

Subtract 2 from both sides to set the equation to 0: \( 2x^2 - 4x + 2 = 0 \). Divide every term by 2 to simplify: \( x^2 - 2x + 1 = 0 \).
04

Solve the Quadratic

Recognize \( x^2 - 2x + 1 = (x-1)^2 \). Thus, \((x-1)^2 = 0\) which means \( x = 1 \). This is the only solution for \( x \).
05

Find the Corresponding \( y \) Value

Substitute \( x = 1 \) back into \( y = 2 - x \) to find \( y \). So, \( y = 2 - 1 = 1 \).
06

Verify the Solution

Check the solution \( (x, y) = (1, 1) \) against both original equations. For \( x + y = 2 \), \( 1 + 1 = 2 \) which is correct. For \( x^2 + y^2 = 2 \), \( 1^2 + 1^2 = 2 \) confirming the solution is correct.

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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 consists of two or more equations that share the same variables. Solving a system of equations requires finding values for the variables that satisfy all the equations simultaneously. In this exercise, we deal with a system of two equations:
  • Equation 1: \( x + y = 2 \)
  • Equation 2: \( x^2 + y^2 = 2 \)
The goal is to find values for \( x \) and \( y \) that make both equations true at the same time. We use the substitution method, a powerful technique for solving such systems.
To apply this method correctly, solve one equation for one variable and substitute it into the other equation. This reduces the two equations to a single equation with one variable, making it easier to solve.
Quadratic Equation
In the process of solving our system, we encounter a quadratic equation. Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In this case, solving for \( x \) after substitution yields a simplified quadratic:
  • \( 2x^2 - 4x + 4 = 2 \), which simplifies to \( x^2 - 2x + 1 = 0 \)
Quadratic equations can often be factored, as is the case here: \((x-1)^2 = 0\). By recognizing this as a perfect square, it's clear that the equation has a single solution, \( x = 1 \). This is known as the "double root" because it satisfies the equation even when expressed as \( (x-1)(x-1) = 0 \). This feature is crucial in finding the unique solutions to the original problem.
Solution Verification
Once a solution is found, verification is crucial to ensure accuracy and correctness. For our solution, \( (x, y) = (1, 1) \), we substitute these values back into the original equations to verify:
  • For \( x + y = 2 \), substituting gives \( 1 + 1 = 2 \), which confirms the solution satisfies the first equation.
  • For \( x^2 + y^2 = 2 \), substituting gives \( 1^2 + 1^2 = 2 \), also confirming the solution satisfies the second equation.
Verification ensures no mistakes were made in the steps of the solution process. Always perform this step to validate your answers, as it reinforces understanding and accuracy.

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Most popular questions from this chapter

In calculus, when finding the derivative of equations in two variables, we typically use implicit differentiation. A more direct approach is used when an equation can be solved for one variable in terms of the other variable. In Exercises \(77-80\), solve each equation for \(y\) in terms of \(x\). $$3 x y=-x^{3} y^{2}, y<0$$

Graph the second-degree equation. (Hint: Transform the equation into an equation that contains no \(x y\) -term.) $$37 x^{2}-42 \sqrt{3} x y+79 y^{2}-400=0$$

Determine whether each statement is true or false. The graph of the equation \(x^{2}+k x y+9 y^{2}=5,\) where \(k\) is any positive constant less than \(6,\) is an ellipse.

In calculus, some operations can be simplified by using parametric equations. Finding the points of intersection (if they exist) of two curves given by parametric equations is a standard procedure. In Exercises \(71-74\), find the points of intersection of the given curves given \(s\) and \(t\) are any real numbers. Curve I: \(x=t, y=t^{2}-1\) Curve II: \(x=s+1, y=4-s\)

A line and a parabola have at most two points of intersection. A line and a cubic function have at most three points of intersection. What is the greatest number of points of intersection that a line and an \(n\) th-degree polynomial can have?
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