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High School Algebra II: Homework Help Resource26 chapters | 280 lessons | 1 flashcard set

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Instructor:
*Mia Primas*

Mia has taught math and science and has a Master's Degree in Secondary Teaching.

Most word problems require that you find the value of one variable. But what if there is more than one unknown variable? In this lesson, you will use systems of equations and solve word problems that have more than one unknown value.

Systems of equations have two or more equations with two or more variables that are solved simultaneously. The goal is to find a value for each variable that satisfies the equations. In the following examples, we will create systems of equations based on the word problems. We will then use substitution and elimination strategies to solve them.

- Example 1: My younger cousins worked part-time jobs over the summer to save money for a car. The job paid different rates for working weekdays and weekends. During the first week, one cousin worked 22 hours during the week and one hour on the weekend, earning $232. Her sister earned $270 by working 15 hours during the week and 10 hours on the weekend. What was the hourly pay rate for working weekdays and weekends?

When solving word problems, it is important to understand what you are asked to find. This problem is asking us to find the hourly pay rate for weekdays and the hourly pay rate for weekends. We will need to use different variables to represent each unknown. Let's use *x* for the weekday rate and *y* for the weekend rate. Now we can use the other information from the word problem to create equations that we will use to solve for the variables.

One cousin worked 22 weekday hours and one weekend hour for a total of $232. We can represent this with the equation 22*x* + *y* = 232. Using the pay for the other cousin, we get the equation 15*x* + 10*y* = 270. Since there are two equations with the same two variables, we can set them up as a system of equations and solve them simultaneously.

The system of equations can be solved using the substitution method. This involves using an expression from one equation to substitute for one of the variables in the other equation. The first equation can be solved for *y* by subtracting 22*x* from both sides of the equation. This leaves us with *y*=232 - 22*x*. Since *y* is equal to the expression 232 - 22*x*, we can substitute the expression for *y* in the second equation. The equation becomes 15*x* + 10(232-22*x*) = 270. We now have one equation with one variable to solve.

We now know that the hourly pay rate for weekdays is ten dollars. By substituting ten for *x* in one of the equations, we can solve for *y*. Using the revised version of the first equation, we get *y* = 232 - 22(10) = 12.

Now that the values of both variables are known, the word problem is solved. The hourly pay rate is ten dollars for weekdays and twelve dollars for weekends.

- Example 2: A high school soccer team held a fundraiser to raise money for new uniforms. They sold cookies and flavored popcorn. On the first day, they sold 44 bags of popcorn and 32 boxes of cookies for a total of $97. On the second day, they raised $179 by selling 68 bags of popcorn and 64 boxes of cookies. How much did they charge for each cookie and bag of popcorn?

In this problem, we are asked to find two things: the price for a box of cookies and the price for a bag of popcorn. We can use *c* to represent the price for a box of cookies and *p* to represent the price for a bag of popcorn. Next we can use the information in the word problem to set up equations that will help us find the values of the variables.

The first day, the soccer team sold 44 bags of popcorn and 32 boxes of cookies for a total of $97. We can represent this with the equation 44*p*+32*c*=97. On the second day, they sold 68 bags of popcorn and 64 boxes of cookies for a total of $179. This can be represented as 68*p*+64*c*=179. Putting these two equations together creates a system of equations.

We can solve the system of equations using the elimination method. Notice that the coefficient of *c* in the second equation is two times the coefficient of *c* in the first equation (64 = 2 x 32). If we multiply the first equation by negative two, the coefficient of *c* will become -64, which is the additive inverse of 64. This ensures that when we add the two equations together the sum of the *c* terms will be zero, thereby eliminating the variable *c* from both equations.

After we add the equations together, we are left with one equation to solve, -20*p* = -15. We can solve for *p* by dividing both sides of the equation by -20. We find that the cost of one bag of popcorn is $0.75.

Now that we have solved for one variable, we can use its value to help us solve for the other value. By substituting 0.75 for *p* in one of the equations in the system of equation, will can find *c*.

This tells us that a box of cookies costs $2.00, and a bag of popcorn costs $0.75.

Systems of equations can be used to solve word problems that have more than one unknown. If there are two unknowns, there should be two equations written based on the information in the word problem. Each equation should have the same two variables. The system of equations can be solved using methods such as substitution and elimination. The problem is solved when values for both unknown variables have been found.

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High School Algebra II: Homework Help Resource26 chapters | 280 lessons | 1 flashcard set

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