This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.

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Solutions to systems of equations come in three types:

**No solutions**. When this happens, we say the system is**inconsistent**. Below is an example of an inconsistent system of equations:\[ x + y = 2 \] \[ x + y = 4 \] There is no solution to this system of equations because you can not find two numbers that add to make \(2\) and \(4\) simultaneously. If you were to proceed with attempting to solve the system, you end up with a contradictory statement - for example, \( 0 = 2 \), or \(4 = 2\).

**Unique solutions**. When this happens, we say the system is consistent. Below is an example of a consistent system of equations with a unique solution: \[ x + y = 4\] \[ x - y = 2\]. Only \(x = 3\) and \(y = 1\) satisfy both equations so this is the unique solution.**Infinitely many solutions**. When this happens, we also say the system is consistent. Below is an example of a consistent system of equations with a infinitely many solutions: \[ 2x - 2y = 8\] \[ -x + y = 4\]. In this case both equations are essentially the same equation (you can see this by multiplying across the second equation by \(-2\)).

These three cases can be represented graphically as follows:

- No Solution (inconsistent system): Parallel lines with no point of intersection (never meet).
- One Unique Solution (consistent system): Lines meet at only one point.
- Infinitely many solutions (consistent system): Coincident lines which meet at all points.

We can also solve systems of linear equations algebraically by substitution.

We use the following system of equations as an example:

`Example`

\(2x+4y=12\) \((1)\)

\(6x-2y=8\) \((2)\)

We first isolate for \(x\) in equation \((1)\) so we can use it in the substitution step:

\(2x+4y=12\)

\(\implies2x=12-4y\)

\(\implies x=\frac{12-4y}{2}\)

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

\(\implies x=6-2y\) \((*)\)

Then, we can substitute the equation \((*)\) into equation \((2)\) to find \(y\):

\(6x-2y=8\)

\(\implies 6(6-2y)-2y=8\)

\(\implies 36-12y-2y=8\)

\(\implies 36-8=12y+2y\)

\(\implies 28=14y\)

\(\implies y=2\)

Now, we can substitute \(y=2\) into the equation \((*)\) to solve for \(x\):

\(x=6-2y\)

\(\implies x=6-2(2)\)

\(\implies x=6-4\)

\(\implies x=2\)

So the unique solution to this system is:

\(x=2\)

\(y=2\)

Another way to solve equations is to use elimination. Here, we aim to 'eliminate' one variable to solve for the other, then substation our solution into our original equations to find the other variable. The steps are as follows:

- Write the equations beneath one another so that the like terms are lined up.
- Multiple all the terms of one equation (or both if necessary) so that the coefficients of one variable are negatives of one another (ex. \(3x\) in one equation and \(-3x\) in the other).
- Add the two equations to get one equation with one unknown, eliminating the other one.
- Solve for the remaining unknown variable.
- Substitute the value found in step 4 into one of the original equations to solve for the other variable.

Be sure to double-check both of your values in the other equation.

Let's see these steps in action!

`Example`

Solve this system of equations using the elimination method:

\(3x+2y=3\) \((1)\)

\(9x-5y=-24\) \((2)\)

**Step 1: **This is already completed as the x and y terms are lined up.

**Step 2: **Let's let the x terms be negatives of one another. Multiply \((1)\) by \(-3\):

\((-3)(3x+2y)=(-3)(3) = -9x-6y=-9\) \((*)\)

**Step 3: **We now add \((*)\) and \((2)\):

\begin{align} -9x - 6y &= -9 \qquad &\textcolor{blue}{(*)} \\ + \qquad 9x - 5y &= -24 \qquad &\textcolor{blue}{(2)} \\ \text{_________}&\text{_______} \\ -11y &= -33 \qquad &\textcolor{blue}{(**)} \end{align}

**Step 4: **We can solve this equation for y:

\(-11y = -33\)

\(\implies y=3\)

**Step 5: **Substitute \(y=3\) into \((1)\) to solve for x:

\(3x+2y=3\)

\(\implies 3x + 2(3) = 3\)

\(\implies 3x + 6 = 3\)

\(\implies 3x = -3\)

\(\implies x = -1\)

**Step 6:** We now have \(x=-1\) and \(y=3\). We double-check that this is our solution by substituting these values into \((2)\):

\(9x-5y = 9(-1) - 5(3) = -9-15 = -24\)

Thus, we have our solution.

You can also solve systems of linear equations using graphs. The solution to a system of linear equations is where the two functions are equivalent. Thus, we can graph the equations and determine the point of intersection. To do this, write each equation while isolating for y to get the equations in slope-intercept form. Then, use this information to graph your functions to find the point of intersection.

Let's look at an example:

`Example`

Solve this system of equations by graphing the equations:

\(6x-3y=-15\) \((1)\)

\(-5x+2y=11\) \((2)\)

`Solution`

First, rearrange the equations and put them in the form \(y=mx+b\).

Equation 1:

\( 6x-3y = -15 \implies -3y = -15-6x \implies y=2x+5\)

Equation 2:

\(-5x+2y = -11 \implies 2y = 11+5x \implies y=\frac{5}{2}x+\frac{11}{2} \)

Now, we can graph these two equations

For equation 1, we have \(y=2x+5\). We know that our y-intercept is \((0, 5)\). Using \(m=\frac{2}{1}\) can find another point on \(y\) by adding 1 to our x-coordinate and 2 to our y-coordinate of the y-intercept to get \((1, 7)\). We can use these two points to graph our equation. |
For equation 2, we have \(y = \frac{5}{2}x+\frac{11}{2}\). Similar to equation 1, we know that our y-intercept is \((0, \frac{11}{2})\). Using \(m=\frac{5}{2}\) can find another point on \(y\) by subtracting 2 from our x-coordinate and 5 from our y-coordinate of the y-intercept to get \((-2, \frac{1}{2})\). We can use these two points to graph our equation. |

If we graph these equations of the same axes, we can see where the point of intersection is

Thus, the solution to this system of equations is \((-1, 3)\)

- Last Updated: Mar 18, 2024 3:47 PM
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