So, again, now we have three equations and three unknowns (variables).
We’ll learn later how to put these in our calculator to easily solve using matrices (see the Matrices and Solving Systems with Matrices section), but for now we need to first use two of the equations to eliminate one of the variables, and then use two other equations to eliminate the same variable: Now this gets more difficult to solve, but remember that in “real life”, there are computers to do all this work!
Let’s let \(j=\) the number of pair of jeans, \(d=\) the number of dresses, and \(s=\) the number of pairs of shoes we should buy.
So far we’ll have the following equations: \(\displaystyle \beginj d s=10\text\\25j \text50d \,20s=260\end\) We’ll need another equation, since for three variables, we need three equations (otherwise, we’d theoretically have infinite ways to solve the problem).
Remember that when you graph a line, you see all the different coordinates (or \(x/y\) combinations) that make the equation work.
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In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later).Note that we solve Algebra Word Problems without Systems here, and we solve systems using matrices in the Matrices and Solving Systems with Matrices section here.“Systems of equations” just means that we are dealing with more than one equation and variable.This will help us decide what variables (unknowns) to use.What we want to know is how many pairs of jeans we want to buy (let’s say “\(j\)”) and how many dresses we want to buy (let’s say “\(d\)”).Then we add the two equations to get “\(0j\)” and eliminate the “\(j\)” variable (thus, the name “linear elimination”). Now that we get \(d=2\), we can plug in that value in the either original equation (use the easiest! We then get the second set of equations to add, and the \(y\)’s are eliminated. Now we can plug in that value in either original equation (use the easiest! Sometimes, however, there are no solutions (when lines are parallel) or an infinite number of solutions (when the two lines are actually the same line, and one is just a “multiple” of the other) to a set of equations.When there is at least one solution, the equations are consistent equations, since they have a solution.The easiest way for the second equation would be the intercept method; when we put for the “\(d\)” intercept.We can do this for the first equation too, or just solve for “\(d\)”.So the points of intersections satisfy both equations simultaneously.We’ll need to put these equations into the \(y=mx b\) (\(d=mj b\)) format, by solving for the \(d\) (which is like the \(y\)): First of all, to graph, we had to either solve for the “\(y\)” value (“\(d\)” in our case) like we did above, or use the cover-up, or intercept method.