Problem 3 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{aligned} y^{2} &=1-x \\ x+2 y &=1 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solutions are (1, 0) and (-3, 2).

Step 1: Solve the Linear Equation for x

The second equation is linear, so we can solve for one variable. Let's solve for \( x \) in the equation \( x + 2y = 1 \).Subtract \( 2y \) from both sides: \[ x = 1 - 2y \]

Step 2: Substitute x in the Quadratic Equation

Now, substitute \( x = 1 - 2y \) into the first equation \( y^2 = 1 - x \).This gives us:\[ y^2 = 1 - (1 - 2y) \]Simplify the equation:\[ y^2 = 2y \]

Step 3: Rearrange and Solve the Quadratic Equation for y

Rearrange the equation \( y^2 = 2y \) to form a standard quadratic equation:\[ y^2 - 2y = 0 \]Factor the equation:\[ y(y - 2) = 0 \]Set each factor equal to zero and solve for \( y \):\[ y = 0 \] or \[ y = 2 \]

Step 4: Find Corresponding x Values

Now, use the values of \( y \) to find \( x \) from \( x = 1 - 2y \).For \( y = 0 \): \[ x = 1 - 2(0) = 1 \]For \( y = 2 \): \[ x = 1 - 2(2) = -3 \]

Step 5: Write the Solution

The solutions to the system of equations are:\( (x, y) = (1, 0) \) and \( (x, y) = (-3, 2) \).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve systems of equations, often involving both linear and quadratic equations. This method requires you to solve one equation for one variable and then substitute that expression into the other equation.

• First, choose the simpler equation of the two — typically the linear equation — and solve for one variable. In this exercise, we solve for \( x \) from the equation \( x + 2y = 1 \), resulting in \( x = 1 - 2y \).
• Next, replace the solved variable in the second equation. By plugging in the expression for \( x \) (from the linear equation), into the quadratic equation \( y^2 = 1 - x \), we substitute \( x \) and transform it into one variable: \( y^2 = 1 - (1 - 2y) \).

This substitution simplifies the problem, allowing for easier manipulation and solution of the equations.

Quadratic Equation

Quadratic equations appear frequently in mathematical problems and are characterized by the presence of the square term \( x^2 \). In our exercise, we encounter a quadratic equation after substituting for \( x \): \( y^2 = 2y \).

• A standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). The aim is to transform any quadratic into this format for easier solving.
• In our case, rearranging the equation \( y^2 = 2y \) gives us \( y^2 - 2y = 0 \), a simplified quadratic expression.
• Factoring helps in finding the solutions easily. Here, the equation factors to \( y(y - 2) = 0 \), revealing the potential solutions for \( y \).

Quadratic equations often yield two solutions, as is evident from the factorization step.

Linear Equation

Linear equations are the foundation for many algebraic processes and constitute one of the two key components of systems of equations. These equations involve variables raised only to the first power and usually take the form \( ax + by = c \).

• In this problem, the linear equation given is \( x + 2y = 1 \). It's simple and can be quickly manipulated to solve for one of the variables. We solve for \( x \), deriving \( x = 1 - 2y \), which we then use for substitution.
• Linear equations often guide the way in systems of equations since they provide clear, direct pathways to finding variable values.

Linear equations are crucial because they provide the first step in applying the substitution method, setting the stage for solving more complex problems.

Solution of Equations

Finding the solution to a system of equations involves determining the set of values for the variables that satisfy all the equations simultaneously. In our exercise, we've used substitution and basic algebraic techniques to find these solutions.

• After substituting and rearranging, the solutions for \( y \) from the quadratic equation \( y^2 - 2y = 0 \) are \( y = 0 \) and \( y = 2 \).
• Using these \( y \)-values, we computed the corresponding \( x \)-values from the equation \( x = 1 - 2y \). For \( y = 0 \), \( x = 1 \), and for \( y = 2 \), \( x = -3 \).
• Thus, the solutions to the system are \((x, y) = (1, 0)\) and \((x, y) = (-3, 2)\).

This set of solutions is where the two original equations intersect, satisfying both conditions set by the linear and quadratic equations.