Solving Systems By Substitution Word Problems

Solving Systems By Substitution Word Problems-84
When equations have no solutions, they are called inconsistent equations, since we can never get a solution.Here are graphs of inconsistent and dependent equations that were created on the graphing calculator: Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns with two equations.

When equations have no solutions, they are called inconsistent equations, since we can never get a solution.Here are graphs of inconsistent and dependent equations that were created on the graphing calculator: Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns with two equations.

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“Systems of equations” just means that we are dealing with more than one equation and variable.In this type of problem, you would also have/need something like this: .Now, since we have the same number of equations as variables, we can potentially get one solution for the system.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 “\(d\)”.When there is only one solution, the system is called independent, since they cross at only one point.When equations have infinite solutions, they are the same equation, are consistent, and are called dependent or coincident (think of one just sitting on top of the other).Now let’s see why we can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below.Remember these are because of the Additive Property of Equality, Subtraction Property of Equality, Multiplicative Property of Equality, and Division Property of Equality: \(\displaystyle \begin\color\\\,\left( \right)\left( \right)=\left( \right)6\text\\,\,\,\,-25j-25d\,=-150\,\\,\,\,\,\,\underline\text\\,\,\,0j 25d=\,50\\25d\,=\,50\d=2\\d j\,\,=\,\,6\\,2 j=6\j=4\end\). \(\displaystyle \begin\color\,\,\,\,\,\,\,\text-3\\color\text\,\,\,\,\,\,\,\text5\end\) \(\displaystyle \begin-6x-15y=3\,\\,\underline\text\\,29x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=58\\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\2(2) 5y=-1\\,\,\,\,\,\,4 5y=-1\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5y=-5\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-1\end\) ). In the example above, we found one unique solution to the set of equations.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!

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