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Question 601434: Hi can you please help me figure this problem. I am solving for x and y. the equation is {(3x+4y=48,x-y=2) this is how the teacher wrote it out on my assignment. The answers given were x=40 and y=42 I need to understand how these numbers were gotten and the book is no help. I am visual and there are no examples close to this problem.
Found 2 solutions by Alan3354, bucky:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! the equation is {(3x+4y=48,x-y=2) this is how the teacher wrote it out on my assignment. The answers given were x=40 and y=42 I need to understand how these numbers were gotten and the book is no help. I am visual and there are no examples close to this problem.
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Firstly, that's not an equation. It's 2 equations, called a system of equations.
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2nd, 40 & 42 are not the answers.
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3x+4y=48
x-y=2 --> x = y+2
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Sub for x in the 1st eqn
3(y+2) + 4y = 48
6y+6 + 4y = 48
7y = 42
y = 6
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x = y+2
x = 8
Answer by bucky(2189)  (Show Source):
You can put this solution on YOUR website! What you were given is two equations as follows:
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3x + 4y = 48 and
x - y =2
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Here's something that will help you. If you have two unknowns (in this case the unknowns are x and y) you will need to have two independent equations in order to determine if a solution can be found for both x and y.
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One of the ways you can solve for the unknowns in these two equations is called variable elimination. We want to make is so one of the variable terms is equal in both the equations. How do we do this? We can multiply both sides of one of the equations by a constant so that one of its terms is equal to the corresponding term in the other equation. (In some problems you may need to multiply both equations by different constants. In this problem, you only need to multiply one of the equations.)
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Let's multiply the second (bottom) equation by 3. That means you will multiply all terms on both sides of the second equation by 3. When you do that the second equation becomes 3x - 3y = 6
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Now write the equations one above the other again, but this time the new bottom equation will be 3x - 3y = 6 as shown below:
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3x + 4y = 48
3x - 3y = 6
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Now subtract the two equations in vertical columns. In the first column you have 3x minus the 3x below it. The answer is zero. In the next column you have 4y minus -3y and by the rules of subtraction, this results in 7y. Finally on the right side you subtract 6 from 48 and the answer is 42. So you have now combined the two equations into a single equation:
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0 + 7y = 42
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Dropping the zero reduces the equation to:
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7y = 42
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and by dividing both sides by 7 you get
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y = 6
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Taking this answer back to either of the original equations and substituting it for y will allow you to solve for x. Let's take it back to the original equation:
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x - y = 2
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Substituting 6 for y gives you:
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x - 6 = 2
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add 6 to both sides to get rid of the -6 on the left side and you have:
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x = 8
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These two answers (y = 6 and x = 8) are correct for the two equations that you gave. And since they do not agree with the answers you were given I suspect that you made a mistake with a sign in one of the equations. For example, if the second equation were supposed to be:
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x + y = 2
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instead of the x - y = 2 that you listed, the equation pair would have been:
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3x + 4y = 48 and
x + y = 2
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Following the same process as above, multiply the bottom equation by 3 and the equation pair then becomes:
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3x + 4y = 48 and
3x + 3y = 6
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Now subtract vertically in columns and you get
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0 + y = 42
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Which simplifies to y = 42
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Returning this answer to the original second equation:
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x + y = 2
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and substituting y = 42 makes it become:
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x + 42 = 2
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Subtracting 42 from both sides results in:
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x = -40
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So this results in y = 42 and x = -40
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Closer, but still not exactly the answer you were given.
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I'm not sure where the error is (maybe your teacher made an error in sign) but there obviously is one some place. Check your problem and see if you can see one or more errors in sign. Then follow the above method to solve the corrected equation or equations to get the teacher's answer. If you can't find any error in the signs of the equations that you were given, try plugging in the teacher's answers of x = 40 and y = 42. You will see that they don't work. For example the first equation (when you substitute 40 for x and 42 for y results in:
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3*(40) + 4*(42) = 48
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Doing the multiplication gives:
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120 + 168 = 48
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Obviously something is wrong because the left side does not equal the right side.
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and substituting 40 for x and 42 for y in the second equation:
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x - y = 2
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results in:
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40 - 42 = 2
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The left side is -2 and the right side is + 2, so that doesn't work either.
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Like I said previously, look for one or more errors in sign in the original problem.
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Just to make sure, let's substitute the answers we got of y = 6 and x = 8 into the two equations you were given to make sure that they work. First:
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3x + 4y = 48
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3*8 + 4*6 = 48
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24 + 24 = 48
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Well, that works because the left side does equal the right side. Let's also do the same for the second equation:
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x - y = 2
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8 - 6 = 2
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That also works. Either the equations the teacher gave you, or the answers you were given were wrong.
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I hope this helps you struggle your way through this problem. If not, ask your teacher to write out the equations again or to recalculate the answers because something is amiss here. Good luck.
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