Тема: Need Math Help

Now we'll move on to some word problems that involve finding where two lines intersect—in other words, solving a system of equations. We'll start with.

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we''''''''''''''''''''''''''''''''ve practiced turning words into linear equations, let''''''''''''''''''''''''''''''''s actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she''''''''''''''''''''''''''''''''ll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

Lots of word problems can be solved with systems of linear equations. However, before we bother with those, let''''s look at some word problems that describe single lines, and we''''re not referring to orderly rows of petrified eighth graders on their way into a school dance.

The amount Jenna makes depends on how many items she sells, so our independent variable x should be the number of items she sells, and the dependent variable y should be the amount she''''s paid. On the graph, the horizontal axis will represent the number of items Jenna sells during one hour, and the vertical axis will represent the amount she gets paid during that hour. By the way, \$3 is quite a commission rate, considering that most of the shop''''s inventory consists of cheap polyester scarves that sell for \$10 a pop. Her dad must run the place.

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations. Then we moved onto solving systems using the Substitution Method. In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we''''''''''''''''ve practiced turning words into linear equations, let''''''''''''''''s actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she''''''''''''''''ll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

Lots of word problems can be solved with systems of linear equations. However, before we bother with those, let''s look at some word problems that describe single lines, and we''re not referring to orderly rows of petrified eighth graders on their way into a school dance.

The amount Jenna makes depends on how many items she sells, so our independent variable x should be the number of items she sells, and the dependent variable y should be the amount she''s paid. On the graph, the horizontal axis will represent the number of items Jenna sells during one hour, and the vertical axis will represent the amount she gets paid during that hour. By the way, \$3 is quite a commission rate, considering that most of the shop''s inventory consists of cheap polyester scarves that sell for \$10 a pop. Her dad must run the place.

x-y = 4 add y to both sides. x - y = 4 +y +y _________ x = y+ 4 use the second equation slove for y x + y = 12 ( y + 4 ) + y = 12 2y + 4 = 12 - 4 - 4 __________ 2y = 8 y= 4 now solve for x x - y = 4 x -( 4) = 4 + 4 + 4 ________ x = 8

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we've practiced turning words into linear equations, let's actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she'll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''ve practiced turning words into linear equations, let''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''s actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''ll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

Lots of word problems can be solved with systems of linear equations. However, before we bother with those, let''''''''s look at some word problems that describe single lines, and we''''''''re not referring to orderly rows of petrified eighth graders on their way into a school dance.

The amount Jenna makes depends on how many items she sells, so our independent variable x should be the number of items she sells, and the dependent variable y should be the amount she''''''''s paid. On the graph, the horizontal axis will represent the number of items Jenna sells during one hour, and the vertical axis will represent the amount she gets paid during that hour. By the way, \$3 is quite a commission rate, considering that most of the shop''''''''s inventory consists of cheap polyester scarves that sell for \$10 a pop. Her dad must run the place.

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations. Then we moved onto solving systems using the Substitution Method. In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

As you get further into Algebra 1, you will find that the real world problems become more complex. They have more questions to be answered and require more steps to find the solution.

But. don't let that intimidate you! You have all the skills that you need to solve these problems. Take one step at a time and think about what you need in order to answer the question. Read through my example very carefully, and study how I performed each step. Pay careful attention to the key words (highlighted words) and how each inequality was written based on the problem.

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we''ve practiced turning words into linear equations, let''s actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she''ll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we''''''''ve practiced turning words into linear equations, let''''''''s actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she''''''''ll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

Lots of word problems can be solved with systems of linear equations. However, before we bother with those, let's look at some word problems that describe single lines, and we're not referring to orderly rows of petrified eighth graders on their way into a school dance.

The amount Jenna makes depends on how many items she sells, so our independent variable x should be the number of items she sells, and the dependent variable y should be the amount she's paid. On the graph, the horizontal axis will represent the number of items Jenna sells during one hour, and the vertical axis will represent the amount she gets paid during that hour. By the way, \$3 is quite a commission rate, considering that most of the shop's inventory consists of cheap polyester scarves that sell for \$10 a pop. Her dad must run the place.

Others have alread posted the solutions for the linear systems. For the factoring problems, first look for the factors of the coefficient of the first term. In the first problem, the only factors of 5 are 1 and 5. So we must be looking at (5x)(1x) to get 5x^2.Then look at factors of the last term. Here, the possible factors are 1x4 or 2x2. Next, look at the middle term and see what combination of the possible first and last could possibly give you the middle term. Here it is 5x+4x. It is clear from this that the last term must factor to 1 and 4, not 2 and 2. So the only reasonable factoring would be (5x+4) (x+1). Use the same procedure for the later problem: the factors of 49 are 1x49 or 7x7. The factors of 1 are 1x1. 14x = 7x+7x, so we get (7x+1)(7x+1). For solving equations, the object is always to transform the equation so that you have a single variable alone on one side of the equation. x-3x^2=0 subtract x from both sides 3x^2=-x factor an x out of both sides x(3x)=x(-1) divide both sides by x 3x=-1 divide both sides by 3 x=-1/3 _________________________ 6x^2-11x=10 subtract 10 from both sides 6x^2-11x-10=0 factor (-3x-2)(-2x+5)=0 solve each factor independently -3x-2=0 -3x=2 x=-2/3 -2x+5=0 -2x=-5 x=5/2

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we''''ve practiced turning words into linear equations, let''''s actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she''''ll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

“Systems of equations” just means that we are dealing with more than one equation and variable.  So far, we’ve basically just played around with the equation for a line, which is y = m x + b.

But let’s say we have the following situation.  You’re going to the mall with your friends and you have \$200 to spend from your recent birthday money. You discover a store that has all jeans for \$25 and all dresses for \$50.  You really, really want to take home 6 items of clothing because you “need” that many new things.

Now that we''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''ve practiced turning words into linear equations, let''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''s actually solve a couple of word problems. This is usually a three-step process:

Jenna works at a retail shop. Yes, she still works there, even after all her thievery, but she''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''ll tell you it has nothing to do with her old man owning the joint. She still makes \$10 per hour, plus \$3 for each item she sells.

Lots of word problems can be solved with systems of linear equations. However, before we bother with those, let''''''''''''''''s look at some word problems that describe single lines, and we''''''''''''''''re not referring to orderly rows of petrified eighth graders on their way into a school dance.

The amount Jenna makes depends on how many items she sells, so our independent variable x should be the number of items she sells, and the dependent variable y should be the amount she''''''''''''''''s paid. On the graph, the horizontal axis will represent the number of items Jenna sells during one hour, and the vertical axis will represent the amount she gets paid during that hour. By the way, \$3 is quite a commission rate, considering that most of the shop''''''''''''''''s inventory consists of cheap polyester scarves that sell for \$10 a pop. Her dad must run the place.

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations. Then we moved onto solving systems using the Substitution Method. In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

As you get further into Algebra 1, you will find that the real world problems become more complex. They have more questions to be answered and require more steps to find the solution.

But. don''t let that intimidate you! You have all the skills that you need to solve these problems. Take one step at a time and think about what you need in order to answer the question. Read through my example very carefully, and study how I performed each step. Pay careful attention to the key words (highlighted words) and how each inequality was written based on the problem.

question #a million: complete form of campers (x) and counselors (y) = 3 hundred for this reason the equation is x + y = 3 hundred. question #2: distinction between form of campers (x) and form of counselors (y) = 280. for this reason the equation is x - y = 280. question #3: The linear equipment includes the x and y equations grouped mutually. The linear equipment subsequently is: x + y = 3 hundred x - y = 280 It did not ask this, yet i pass to clean up the equipment. x + y = 3 hundred y = 3 hundred - x x - y = 280 x - (3 hundred - x) = 280 x - 3 hundred + x = 280 2x - 3 hundred = 280 2x = 580 x = 290 for this reason 290 - y = 280 y = 10 for this reason, there are 290 campers and 10 counselors.