The first one we used was how to write equations from word problems. To solve that example problem, we used several different math concepts. By substituting y with its value of 10, we would get 0.8(10)=x which would give us a value of 8 for x. Now that we have found the value of the y variable, we can plug it back into the equation, 0.8y=x, to find the value of x. The next step would be to solve this equation for the y variable by combining like terms: 2.8y=28 which would give us y=10 or 10 Spanish questions answered. To do this method, we would substitute these equations for the x and y variables of the third equation which would give us (0.8y)+y+(y+5)=33. Because there are two equations already solved in terms of two variables, 0.8y=x and z=y+5, substitution would be the most efficient method. Since the first equation we found was 0.8y=x, we can see that we only need the y variable to find the value of x, or the amount of math questions that were answered. Looking back at the original question, the goal of this problem is to find how many math questions Logan has answered. The final equation would be x+y+z=33 to represent how Logan answered 33 questions in total. The second equation would be z=y+5 to represent how Logan has answered five more English questions than Spanish ones. The variables x, y, and z will represent the amount of math, Spanish, and English questions that Logan has answered respectively.Īs Logan has answered 0.8 times as many math questions as Spanish ones, the equation to represent this would be 0.8y=x. To do this, we must assign variables to each unknown part of the problem. The first step is to create equations from the word problem. If Logan has answered 33 questions in total, how many math questions has Logan answered? Logan has answered 0.8 as many math questions as Spanish questions, and he’s answered 5 more English questions than Spanish questions. Here’s an example of a problem where solving a system of equations is necessary: The next step would be to use the equation that we created to find the value of the variable and then plug that value back into an original equation to find the remaining variable. Elimination is best used when this is already occurring in the equations, but the equations can also be manipulated into creating common coefficients by either multiplying or dividing equations by a certain number. This can only be done when the coefficients of one variable in both equations are opposites and will cancel each other out once added together. Elimination is adding the equations together in order to create an equation with only one variable. The final step is to substitute the number value that was found in for its corresponding variable in the original equation. Once an expression for the variable is found, substitute or plug in the expression into the other equation where the original variable was to solve for the number value of the next variable. The first step in this method is to solve one of the equations for one variable. Substitution is best used when one of the equations is in terms of one of the variables such as y=2x+4, but the equations can always be manipulated. This is most efficient when the equations are already written in slope-intercept form. The coordinate of this point will give you the values of the variables that you are solving for. To solve a system by graphing, you simply graph the given equations and find the point(s) where they all intersect. There are three methods used to solve systems of equations: graphing, substitution, and elimination. To solve systems of equations, it is helpful for students to have a background understanding of simple algebraic equations, variables, and graphing linear equations. When this happens, the goal for students is to use given information in the equations to solve for all variables. Students encounter these systems of equations when there are multiple ‘unknowns’ - or variables - that have not been given to them yet. Systems of equations are multiple equations that all have a common solution.
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