# Using Linear Systems To Solve Problems

Once one variable is eliminated, it becomes much easier to solve for the other one.Multiplication can be used to set up matching terms in equations before they are combined.

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You can multiply both sides of one of the equations by a number that will result in the coefficient of one of the variables being the opposite of the same variable in the other equation. Notice that the first equation contains the term 4y, and the second equation contains the term y.

If you multiply the second equation by −4, when you add both equations the y variables will add up to 0. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied both equations by different numbers. Adding 4x to both sides of Equation A will not change the value of the equation, but it will not help eliminate either of the variables—you will end up with the rewritten equation 7y = 5 4x.

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 \(d\): \(\displaystyle d=-j 6\).

If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables.

So let’s now use the multiplication property of equality first.

The first trick in problems like this is to figure out what we want to know.

This will help us decide what variables (unknowns) to use.

When using the multiplication method, it is important to multiply all the terms on both sides of the equation—not just the one term you are trying to eliminate.

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.

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