Linear Equations

Up until now, when solving algebra equations, we've mostly worked with equations that only have one unknown. Therefore, we were able to find that unknown value. Equations have more than one unknown, and can not necessarily be solved for a particular value. Instead we need to think of them visually. Let's consider the following equation with two unknowns

\[y = 3x + 2\]

Looking at this graph, we can see this defines a straight line. This is what we mean by a linear equation. Linear equations have the following general form

\[y = mx + c\]

The \(c\) can be thought of as the intersect. This is the point where the line crosses the y-axis. In our example \(c=2\) and the line does cross the y-axis at 2. If \(c\) was negative it would cross the y-axis below the x-axis.

\(m\) is often called the gradient or slope. This is the steepness of the line. A larger slope means a steeper line. A negative slope means the line points in the opposite direction. A slope of 0 is a flat, horizontal line. A slop of infinity is a vertical line.

Let's say that you are given the following two pieces of information. A line has value \(y=-10\) when \(x=1\) and \(y=-2\) when \(x=-1\). Can we make the straight line equation for this line?

First, let's find the slope. Slope can be thought of the change in height over the change in length. So in our case, the height goes from -2 to -10, when the length goes from -1 to 1.

\[\frac{-10 - - 2}{1 - - 1} = \frac{-8}{2} = -4\]

So, we already have \(y = -4x + c\). Now to find the \(c\) we simply use what we've already been given. At \(x=1, y=-10\)

\[-10 = -4 + c\]

\[c = - 6\]

So the full line equation is \(y = -4x - 6\), which is graphed as

Questions

Find the equation of the straight line that passes through the following sets of points. (We often write a point as \((x,y)\))

1. \((1,3), (2,9)\)
2. \((-1, 6), (4, 4)\)
3. \((2, 2), (12, 12)\)
4. \((-13, 1), (25, 1)\)

Simultaneous Equations

We already looked at linear equations. Remember how with one unknown, we could find its value? And with two unknowns and one equation, we couldn't find a unique value, but could graph a range of possible solutions? What about when we have two unkowns and two equations? Imagine we have the following two equations.

\[y = 2x + 2, y = -2x + 3\]

Visually this is

We can think of this as trying to find the values of the point where they cross. Simultaneous equations can only have a real solution if the lines do in fact cross.

We can solve these equations like this.

\[2x + 2 = -2x + 3\]

\[4x = 1\]

\[x = \frac{1}{4}\]

\[y = 2 \frac{1}{4} + 2 = 2.5\]

There are many ways to solve simultaneous equations. The two most common is elimination and substitution.

Consider the equations

\[2x + y = 1, 3x - 2y = 4\]

Notice how one has \(y\) and the other has \(-2y\)? What if we multiplied the first equation through by 2 to give

\[4x + 2y = 2\]

and now we add this to the second equation

\[3x - 2y + 4x + 2y = 2 + 4\]

\[7x = 6\]

\[x = \frac{6}{7}

Now, we can plug this into any equation we like

\[2 \frac{6}{7} + y = 1\]

\[y = \frac{-5}{7}\]

Let's try this again but with a substitution. If we rearrange the first equation into.

\[y = 1 - 2x\]

And substitute it into the other equation

\[3x - 2(1 - 2x) = 4\]

\[3x - 2 + 4x = 4\]

\[7x = 6\]

\[x = \frac{6}{7}\]

\[y = 1 - 2 \frac{6}{7} = \frac{-5}{7}\]

There is no one way to solve a simultaneous equation, and you might need to use a few different techniques to coerce a solution out.

Questions

Find the solutions to the following pairs of simultaneous equations

1. \(5x + 3y = 41, 2x + 3y = 20\)
2. \(5x + y = 11, 3x - y = 9\)
3. \(x + 7y = 64, x + 3y = 28\)
4. \(4x - 4y = 24, x - 4y = 3\)

Let's take what you have learned and solve a very challenging set of simultaneous equations. Solve this set of four equations

\[p + q + r = 0\]

\[s = 2 - 200p\]

\[s = 4 - 300q\]

\[s = -100r\]