Lane ORCCA (2020–2021): Open Resources for Community College Algebra

To start, we want to see whether it will be easier to eliminate \(x\) or \(y\text{.}\) We see that the coefficients of \(x\) in each equation are \(3\) and \(5\text{,}\) and the coefficients of \(y\) are \(-4\) and \(8\text{.}\) Because \(8\) is a multiple of \(4\) and the coefficients already have opposite signs, the \(y\) variable will be easier to eliminate.

To eliminate the \(y\) terms, we will multiply each side of the first equation by \(2\) so that we will have \(-8y\text{.}\) We can call this process scaling the first equation by \(2\text{.}\)

We now have an equivalent system of equations where the \(y\)-terms can be eliminated:

To solve for \(y\text{,}\) we will substitute \(2\) for \(x\) into either of the original equations or the new one. We will use the original first equation, \(3x-4y=2\text{:}\)

Our solution is \(x=2\) and \(y=1\text{.}\) We will check this in both of the original equations:

The solution to this system is \((2,1)\) and the solution set is \(\{(2,1)\}\text{.}\)

Considering the coefficients of \(x\) (\(2\) and \(-3\)) and the coefficients of \(y\) (\(3\) and \(5\)), we see that we cannot eliminate the \(x\) or the \(y\) variable by scaling a single equation. We will need to scale both.

The \(x\)-terms already have opposite signs, so we choose to eliminate \(x\text{.}\) The least common multiple of \(2\) and \(3\) is \(6\text{.}\) We can scale the first equation by \(3\) and the second equation by \(2\) so that the equations have terms \(6x\) and \(-6x\text{,}\) which will cancel when added.

At this point we can add the corresponding sides from the two equations and solve for \(y\text{:}\)

To solve for \(x\text{,}\) we'll replace \(y\) with \(0\) in \(2x+3y=10\text{:}\)

We'll check the system using \(x=5\) and \(y=0\) in each of the original equations:

So the system's solution is \((5,0)\) and the solution set is \(\{(5,0)\}\text{.}\)

To solve the system using elimination, we first need to scale one or both of the equations so that one variable has equal but opposite coefficients in the system. In this case, we will choose to make \(y\) have opposite coefficients because the signs are already opposite for that variable in the system.

To determine how to scale each equation, we first find the least common multiple of the coefficients of \(y\text{.}\) The least common multiple of \(6\) and \(4\) is \(12\text{.}\) So, this is the coefficient we want on each \(y\text{,}\) but with opposite signs of course. Thus, we need to multiply the first equation by \(2\) and the second equation by \(3\text{.}\)

We now have an equivalent system of equations where the \(y\)-terms can be eliminated:

To solve for \(y\text{,}\) we will substitute \(\substitute{1}\) for \(\substitute{x}\) into either of the original equations. We will use the first equation, \(4x-6y=13\text{:}\)

To verify this, we substitute the \(x\) and \(y\) values into both of the original equations.

So the solution is the point \(\left(1, -\frac{3}{2}\right)\) and the solution set is \(\left\{\left(1,-\frac{3}{2}\right)\right\}\text{.}\)

To eliminate the \(x\)-terms, we multiply each term in the first equation by \(-2\text{,}\) and we have:

We might notice that the equations look very similar. Adding the respective sides of the equation, we have:

Both of the variables have been eliminated. Since the statement \(0=0\) is true no matter what \(x\) and \(y\) are, the solution set is infinite. Specifically, you just need any \((x,y)\) satisfying one of the two equations, since the two equations represent the same line. We can write the solution set as \(\{(x,y)\mid 3x+4y=5\}\text{.}\)

To eliminate the \(x\)-terms, we will scale the first equation by \(-5\) and the second by \(2\text{:}\)

Adding the respective sides of the equation, we have:

Both of the variables have been eliminated. In this case, the statement \(0=-37\) is just false, no matter what \(x\) and \(y\) are. So the system has no solution.

Since the variable \(x\) is already isolated here, it is set up nicely to use the substitution method. So that is the method we will use here.

We start by substituting the expression \(3y-8\) in place of \(x\) in the second equation and then solve for \(y\text{:}\)

Next, we can find the value of \(x\) by substituting \(5\) in place of \(y\) in the first equation:

Therefore, the solution to the system is \((7,5)\text{.}\)

First, we can cancel the fractions by using the least common multiple of the denominators in each equation, similarly to the topic of Section 3.3. We have:

We could put convert the first equation into standard form by subtracting \(x\) from both sides, and then use elimination. However, the \(x\)-variable in the first equation has a coefficient of \(1\text{,}\) so the substitution method may be faster. Solving for \(x\) in the first equation we have:

Using the equation where we isolated \(x\) and substituting \(-1\) for \(y\text{,}\) we have:

The solution is \((2,-1)\text{.}\) Checking the solution is left as an exercise.

Since the coefficient of \(x\) (or \(y\)) in the first equation is \(1\text{,}\) we could solve for one of these variables and use substitution to complete the problem. However, some decimal arithmetic would be required. Alternatively, we can scale the equations by the right power of \(10\) to make all the coefficients and constants integers:

Now to set up elimination, scale each equation again to eliminate \(x\text{:}\)

Adding the corresponding sides from the two equations gives

from which we find \(y=0.1\text{.}\)

To solve for \(x\text{,}\) we can use one of the original equations:

Therefore, the solution to the system is \((0.25,0.1)\text{.}\)