Question

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{r} 9 x^{2}-8 x y+4 y^{2}=70 \\ 3 x+2 y=10 \end{array}\right. $$

Short answer

Solutions are \((2, 2)\) and \((1, \frac{7}{2})\).

Step-by-step solution

Solve the Linear Equation for One Variable

First, solve the linear equation for one of the variables. Let's solve the linear equation for y: \[ 3x + 2y = 10 \]Subtract 3x from both sides: \[ 2y = 10 - 3x \]Then, divide both sides by 2 to solve for y: \[ y = 5 - \frac{3}{2} x \]

Substitute Back into the Quadratic Equation

Next, substitute the expression for y back into the quadratic equation: \[ 9x^2 - 8xy + 4y^2 = 70 \]Substitute \( y = 5 - \frac{3}{2} x \) into the equation: \[ 9x^2 - 8x(5 - \frac{3}{2} x) + 4(5 - \frac{3}{2} x)^2 = 70 \]

Simplify the Quadratic Equation

Expand and simplify the substituted quadratic equation: \[ 9x^2 - 40x + 12x^2 + 4(25 - 15x + \frac{9}{4} x^2) = 70 \]Combine like terms: \[ 21x^2 - 40x + 100 - 60x + 9x^2 = 70 \]Further simplifying: \[ 30x^2 - 100x + 100 = 70 \]

Solve the Simplified Quadratic Equation

Subtract 70 from both sides to set the equation to 0: \[ 30x^2 - 100x + 30 = 0 \]Factor out the greatest common factor (GCF): \[ 30(x^2 - \frac{10}{3} x + 1) = 0 \]Solve the factored quadratic equation: \[ x = \frac{5 \pm \sqrt{1}}{3} \]Thus, the solutions for x are: \[ x = 2 \text{ and } x = 1 \]

Find Corresponding y Values

Using the linear equation, find the corresponding y values for each x: For \( x = 2 \): \[ y = 5 - \frac{3}{2} (2) = 2 \]For \( x = 1 \): \[ y = 5 - \frac{3}{2} (1) = \frac{7}{2} \]

Verify the Solutions

Verify the solutions by substituting \((x, y) = (2, 2)\) and \((x, y) = (1, \frac{7}{2})\) back into the original quadratic equation.

Key concepts

Linear Equations

A linear equation represents a relationship between two variables that forms a straight line when plotted on a graph. In this exercise, the linear equation is: \(3x + 2y = 10\).
To solve for one variable, we need to isolate it on one side of the equation. This often involves simple arithmetic such as addition, subtraction, multiplication, or division.
Here, isolating y gives us \(y = 5 - \frac{3}{2}x\). This step allows us to substitute y in the quadratic equation later.

Quadratic Equations

A quadratic equation involves terms up to the second degree and forms a parabola when graphed. The quadratic equation given is: \(9x^2 - 8xy + 4y^2 = 70\).
Quadratic equations can have either zero, one, or two real solutions. They are usually solved using methods like factoring, completing the square, or using the quadratic formula.
In this exercise, we solve the system by first simplifying the quadratic equation after substituting y.

Substitution Method

The substitution method involves solving one equation for one variable and then substituting that solution into another equation.
Here's a breakdown:

• First, solve the linear equation \(3x + 2y = 10\) for y, which gives \(y = 5 - \frac{3}{2}x\).
• Next, substitute this expression for y into the quadratic equation \(9x^2 - 8xy + 4y^2 = 70\).
• This helps us reduce the system to a single variable quadratic equation.

This substitution simplifies the complexity of solving systems of equations.

Factoring

Factoring involves rewriting an equation as a product of its factors. For quadratic equations in the form \(ax^2 + bx + c = 0\), we look for two numbers that multiply to ac and add to b.
In our exercise, we simplify the quadratic equation to \(30x^2 - 100x + 30 = 0\).
We can factor out a greatest common factor (GCF), resulting in \(30(x^2 - \frac{10}{3}x + 1) = 0\).
This further simplifies to solve for x.

Solving Quadratic Equations

There are several methods to solve quadratic equations. In the exercise, after factoring, we solve the simplified quadratic equation: \(30(x^2 - \frac{10}{3}x + 1) = 0\).
This involves setting the factored expression equal to zero: \(x = \frac{5 \pm \sqrt{1}}{3}\), leading to solutions \(x = 1\) and \(x = 2\).

• Next, we find the corresponding y values using \(y = 5 - \frac{3}{2}x\) for each x.
• Finally, we verify the solutions by substituting into the original equations to ensure both x and y satisfy them.

This confirms our solutions are correct.