# Equations With Two Variables In Algebra

These are my notes on equations with two variables in Algebra.

The arithmetic operations of addition, subtraction, multiplication, and division are computed by functions of two inputs. For functions of two inputs, it is common to use x and y as inputs and z as the output. This is expressed as \(z = f(x,y) \).

**Example 1**

Evaluate \( f(3,-4) \) where \( f(x,y) = xy \)

\( f(3,-4) = (3)(-4) \)

\( \boxed{f(3,-4) = -12} \)

**Example 2**

Evaluate \( M(120,5) = \frac{m}{g} \)

\( M(120,5) = \frac{120}{5} \)

\( \boxed{M(120,5) = 24} \)

**Example 3**

Evaluate \( V(0.5,2) = \pi r^{2} h \)

\( V(0.5,2) = \pi(0.5)^{2}(2) \)

\( V(0.5,2) = 0.5 \pi \)

\( \boxed{V(0.5,2) = 1.57} \)

We often use formulas with more than one variable. Sometimes, it is necessary to solve an equation for a variable. For example, to find the radius of a circle that has an area of 50 square inches, we might first solve the equation \( A = \pi r^{2} \) for r.

**Example 4**

Find the radius of a circle that has an area of 50 square inches.

\( A = \pi r^{2} \)

Divide by \(\pi\)

\( \frac{A}{\pi} = r^{2} \)

Apply the square root property

\( r = \sqrt{\frac{A}{\pi}} \)

\( r = \sqrt{\frac{50}{\pi}} \)

\( \boxed{r = 3.99 \text{ inches}} \)

The equation \( P = 2L + 2W \) calculates the perimeter of a rectangle with length and width.

**Example 5**

Solve for L

\( P = 2L + 2W \)

Subtract 2W

\( P - 2W = 2L \)

Divide both sides of the equation by 2

\( \boxed{\frac{P-2W}{2} = L} \)

**Example 6**

Find the length of a rectangle when P=21 and W=3.5 feet

\( L = \frac{P-2W}{2} \)

\( L = \frac{21-2(3.5)}{2} \)

\( \boxed{L = 7} \) feet

A linear equation in two variables can be written as:

\( ax + by = k \)

Where a,b, and k are constants.

Here is an example.

\( f(x,y) = 10 \)

\( g(x,y) = 2 \)

These can be rewritten as:

\( \frac{x+y}{2} = 10 \)

\( x-y = 2 \)

This pair of equations is called a system of linear equations because we are solving more than one linear equation at once. A solution to a system of equations consists of an x-value and a y-value that satisfy both equations simultaneously. The set of all solutions is called the solution set.

The substitution method is often used to solve systems of equations involving two variables.

**Example 7**

Solve:

\( x + y = 60.9 \)

And

\( x - y = 22.1 \)

Begin by solving an equation for a convenient variable. Then substitute the result into the other equation. Solve for y.

\( y = 60.9 -x \)

Now substitute that into the other equation.

\( x - (60.9 - x) = 22.1 \)

Distribute

\( x - 60.9 + x = 22.1 \)

Combine terms

\( 2x = 83 \)

Divide both sides of the equation by 2

\( \boxed{x = 41.5} \)

\( y = 60.9 - 41.5 \)

\( \boxed{y = 19.4} \)

Method of Substitution

- Choose a variable in one of the two equations and solve for that variable
- Substitute the result from step 1 into the other equation and solve for the remaining variable.
- Use the value of the variable from step 2 to determine the value of the other variable. Use the equation you found in step 1.

**Example 8**

Solve:

\( 5x - 2y = -16 \)

\( x + 4y = -1 \)

Solve one of the equations for one of the variable

If given a choice, choose the one without a coefficient as that will make it easier

\( x = -4y - 1 \)

Now substitute that into the first equation and solve for y

\( 5(-4y - 1) - 2y = -16 \)

Use the distributive property

\( -20y -5 -2y = -16 \)

Combine terms

\( -22y -5 = -16 \)

Add 5 to both sides of the equation

\( -22y = -11 \)

\( \boxed{y = \frac{1}{2}} \)

Now substitute y into the equation you used in step 1

We used \( x = -4y -1 \)

\( x = -4(\frac{1}{2}) -1 \)

\( \boxed{x = -3} \)

Our answer is:

\( \boxed{(-3,\frac{1}{2})} \)

The method of substitution can also be used to solve systems of nonlinear equations. In general, a system of nonlinear equations can have any number of solutions.

**Example 9**

Solve:

\( 6x + 2y = 10 \)

\( 2x^{2} - 3y = 11 \)

Begin by solving one of the equations for one of the variables. One possibility is to solve the first equation for y.

\( 6x + 2y = 10 \)

Subtract 6x from each side of the equation

\( 2y = 10 - 6x \)

Divide each side of the equation by 2

\( y = \frac{10-6x}{2} \)

\( y = 5 - 3x \)

Now, substitute \(5-3x\) for y in the second equation and solve the resulting quadratic equation for x.

\( 2x^{2} - 3y = 11 \)

\( 2x^{2} - 3(5-3x) = 11 \)

Distribute

\( 2x^{2} -15 +9x = 11 \)

Subtract 11 from both sides of the equation

\( 2x^{2} + 9x - 26 = 0 \)

Factor

\( (2x+13)(x-2) = 0 \)

\( x = -\frac{13}{2} \text{ or } x = 2 \)

Now find the corresponding y-values for each x-value. From step 1 we know that \(y=5-3x\), so:

\(y = 5 - 3(-\frac{13}{2}) \)

\( y = \frac{49}{2} \)

And

\( y = 5 - 3(2) \)

\( y = -1 \)

The solutions are therefore:

\( \boxed{(-\frac{13}{2},\frac{49}{2}) \text{ and } (2,-1)} \)

A circle with radius r, centered at the origin, has an equation \(x^{2} + y^{2} = r^{2}\). Use the method of substitution to determine the points where the graph of \(y=2x\) intersects the circle when \(r=\sqrt{5}\).

**Example 10**

Solve:

\(x^{2} + y^{2} = r^{2}\)

\(y = 2x\)

Since \(r^{2}=5\)

\( x^{2} + y^{2} = 5 \)

\( y = 2x \)

We already know y, so substitute

\( x^{2} + (2x)^{2} = 5 \)

Square the expression

\( x^{2} + 4x^{2} = 5 \)

Combine terms

\( 5x^{2} = 5 \)

Divide by 5

\( x^{2} = 1 \)

\( x=1 \text{ or } x=-1 \)

Since \(y=2x\), we see that when \(x=1\), \(y=2\). When \(x=-1\), \(y=-2\). The graphs of

\(x^{2} + y^{2} = 5\) and \(y=2x\) intersect at the points \((1,2)\) and \((-1,-2)\).

**Example 11**

Solve:

\( x^{2} + y = 1 \)

\( x^{2} - y = -2 \)

Solve the second equation for y

\( -y = -2 - x^{2} \)

Distribute the negative sign

\( y = x^{2} + 2 \)

Substitute \(x^{2}+2\) for y in the first equation and solve for x

\( x^{2} + (x^{2} + 2) = 1 \)

Combine terms

\( 2x^{2} + 2 = 1 \)

Subtract 2 from each side of the equation

\( 2x^{2} = -1 \)

Because \(2x^{2} \geq 0 \), there are no real solutions if you look at original equation.

**Example 12**

Solve:

\( 2x - 4y = 5 \)

\( -x + 2y = \frac{-5}{2} \)

Solve the second equation for x

\( -x = - 2y - \frac{5}{2} \)

Distribute the negative sign

\( x = 2y + \frac{5}{2} \)

Now substitute \( x=2y+\frac{5}{2} \) in the first equation

\( 2x - 4y = 5 \)

\( 2(2y + \frac{5}{2}) - 4y = 5 \)

Distribute

\( 4y + 5 - 4y = 5 \)

\( 5 = 5 \)

This is an identity that is always true and indicates that there are infinitely many solutions. It means any point on the line represents a solution.

**Example 13**

Determining the dimensions of a cylinder

The volume of a cylindrical container with a radius r and height h is computed by:

\( V(r,h) = \pi r^{2} h \)

The lateral surface area S of the container is computed by \( S(r,h) = 2\pi r h \)

Write a system of equations whose solution is the dimensions for a cylinder with a volume of 38 cubic inches and a lateral surface area of 63 square inches.

Solve the system of equations

The equations \( V(r,h)=38 \) and \( S(r,h) = 63 \) must be satisfied. This results in the following:

\( \pi r^{2}h = 38 \)

\( 2\pi rh = 63 \)

\( h = \frac{38}{\pi r^{2}} \)

\( h = \frac{63}{2\pi r} \)

\( \frac{38}{\pi r^{2}} = \frac{63}{2\pi r} \)

Multiply by the least common denominator

\( 2\pi r^{2} \frac{38}{\pi r^{2}} = 2\pi r^{2} \frac{63}{2\pi r} \)

Simplify

\( 76 = 63r \)

Divide both sides by 63

\( r = \frac{76}{63} \)

\( r = 1.20 \)

\( h = \frac{63}{2\pi r} \)

\( h = \frac{63}{2\pi \frac{76}{63}} \)

\( h = 8.31 \)

A quantity may depend on more than one variable. For example, the volume V of a cylinder is given by \(V = \pi r^{2}h \). We say that V varies jointly as h and the square of r. The constant of variation is \(\pi\).