# Using Linear Systems To Solve Problems Notice that the \(j\) variable is just like the \(x\) variable and the \(d\) variable is just like the \(y\).

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Now, you can always do “guess and check” to see what would work, but you might as well use algebra!

It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

Just as with the substitution method, the elimination method will sometimes eliminate both variables, and you end up with either a true statement or a false statement.

Recall that a false statement means that there is no solution.

This means that the numbers that work for both equations is We can see the two graphs intercept at the point \((4,2)\). It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation.

This means that the numbers that work for both equations is 4 pairs of jeans and 2 dresses! Here is the problem again: Solve for \(d\): \(\displaystyle d=-j 6\).

So if you have a system: x – 6 = −6 and x y = 8, you can add x y to the left side of the first equation and add 8 to the right side of the equation.

And since x y = 8, you are adding the same value to each side of the first equation.

Instead, it would create another equation where both variables are present.

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