System of Equations

Solving Equations

Example 1 — How Many Solutions?

Solution

In this question, notice we are given one equation with two unknown variables. If we isolate for one variable, we will simply get an alternative form of the same equation and not a numeric value.

• Isolating for \(y\rightarrow\quad y=-\dfrac{2}{3}x+4\), or
• Isolating for \(x\rightarrow\quad x=-\dfrac{3}{2}y+6\)

In fact, because we are only given one equation with two variables, we cannot solve for both variables in the equation. In previous lessons we chose a value for \(x\) and could then calculate the corresponding \(y\)-value. As you have seen previously when working with linear relations, there are an infinite number of solutions to this relation, as represented by the graph shown here. For example, \(x=0,\: y=4\), and \(x=-\frac32,\: y=5\) are two such solutions. In fact, every point that lies along the line is a solution.

Solving Systems of Equations

If instead we were asked to consider both of the equations \(2x+3y=12\) and \(x-y=1\) collectively, or as a group, we would be able to solve for both the variables \(x\) and \(y\).

If we are asked to solve the system above, we are being asked to find values of the variables \(x\) and \(y\) that satisfy both of the equations simultaneously. For example, \(x=-3\), \(y=6\) is not a solution to the system of equations above since it does not satisfy the second equation (although it does satisfy the first equation). However, \(x=3\), \(y=2\) is a solution to the system of equations above since these values do satisfy both equations.

When we are asked to solve a linear system of two equations written in terms of two variables \(x\) and \(y\), we are looking for the point \((x,y)\) that both equations share. In other words we are looking for the point of intersection. 

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Example 2

Solution — Part A

For a linear system of equations to have no solution, the lines must be parallel and distinct. The line we are trying to determine will therefore need to have the same slope and a different \(y\)-intercept. It will be easier for us to see the slope and \(y\)-intercept of the given equation if we rearrange it into the form \(y=mx+b\).

\(\begin{align*} 2x+3y&=6\\ 3y&=-2x+6\\ y&=-\dfrac{2}{3}x+2 \end{align*}\)

Therefore, the slope of the second line must be equal to \(-\dfrac{2}{3}\) and the \(y\)-intercept can be any number other than \(2\). One possible solution is \(y=-\dfrac{2}{3}x+10\). We can graph both of these lines to show that they are parallel and distinct, and therefore the system has no solution.

Solution — Part B

For a linear system of equations to have one solution, the slopes must be different. From part a) we know that the slope of the given line is \(-\dfrac{2}{3}\). Therefore, a linear relation with a slope other than \(-\dfrac{2}{3}\) will give us a system of equations with one solution. One possible equation is  \(y=5x+19\). Again, we can graph both of these lines to show that they have one intersection point at \((-3,4)\).  

Solution — Part C

For a linear system of equations to have infinitely many solutions, the second equation must be equivalent to the given equation. We can determine an equivalent representation by multiplying or dividing each term in the equation by the same value.  Let's multiply the original equation by \(-1\).

\(\begin{align*} 2x+3y&=6\\ (-1)(2x)+(-1)(3y)&=(-1)(6)\\ -2x-3y&=-6 \end{align*}\)

We can now graph \(2x+3y=6\) and \(-2x-3y=-6\) and show these are the same line with infinite points of intersection.