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Sholpan [36]
4 years ago
5

Need help with cross multiplying

Mathematics
1 answer:
saw5 [17]4 years ago
4 0
So to solve for this, we need to set up proportional fractions, which I will help show you how to do.
First, if we are given an amount out of a total, we need to put it over x (if we are looking for the total). It looks like this:
12/x, 12 being the given number and x being the total.
If we are given the total but are looking for an amount, put the total at the bottom of the fraction (aka the denominator). It looks like this: x/16, 16 being the total amount and x being the amount out of the total.
We have a total of 40 test problems, so we can put our total at the bottom, x/40.
X is the amount of questions answered correctly (we are looking for x in the question).
We have answered 80% correct, so put 80% over 100 (100 being the total). It should look like this: 80/100.
Now we have our two fractions: x/40 & 80/100.
Set these up as an equation.
x/40 = 80/100.
Now this is where things may get tricky if you don't pay attention.
Multiply the numerator (the top number of a fraction) of x/40 by the denominator (the bottom number of a fraction) of 80/100.
Your product equation should look like this:
x times 100. This will give is 100x. Leave it at that.
Now, multiply the denominator of x/40 (the bottom number of the fraction) by the numerator (the top number of a fraction) of 80/100. It should look like this:
80 x 40. This will give us 3200.
Now set up our products as an equation.
100x = 3200.
To solve for x, divide both sides by 100.
3200/100 = 32.
x = 32.
I hope this helps and has taught you something!
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aniked [119]

Answer:

1) x = 3, y = 2 or (3, 2)

2) x = 0, y = 2 or (0, 2)

6) x = 3, y = – 4 or (3, – 4)

Step-by-step explanation:

I will only provide solutions for questions 1, 3, and 6 in accordance with Brainly's guidelines.

<h2>Question 1:  </h2>

\displaystyle\mathsf{\left \{ {{\:\:Equation\:1:\:y\:=\:-2x\:+\:5} \atop{Equation\:2:\:y\:=\:-x\:+\:2}} \right.  }  

Substitute the value of y in equation 2 into the first equation:

y = -2x + 5

-x + 2 = -2x + 5

Add 2x on both sides:

-x + 2x + 2 = -2x + 2x + 5

x + 2 = 5

Subtract 2 from both sides to isolate x:

x + 2 - 2 = 5 - 2

x = 3

Substitute the value of x into Equation 2 to solve for the value of y:

y = -x + 5

y = -(3) + 5

y = 2

<h3><u>Solution of the given systems of linear equations:</u></h3>

Therefore, the solution to the given systems of linear equations is: x = 3, y = 2 or (3, 2).

<h2>Question 3: </h2>

\displaystyle\mathsf{\left \{ {{Equation\:1:\:3x\:+\:5y=\:10} \atop{Equation\:2:\:y\:=\:-5x\:+\:2}} \right.  }

Substitute the value of y from Equation 2 into Equation 1:

3x + 5y = 10

3x + 5(-5x + 2) = 10

Distribute 5 into the parenthesis:

3x + -25x + 10 = 10

Combine like terms:

-22x + 10 = 10

Subtract 10 from both sides:

-22x + 10 - 10 = 10 - 10

-22x = 0  

Divide both sides by -22 to solve for x:

\displaystyle\mathsf{\frac{-22x}{-22}\:=\:\frac{0}{-22}}

x = 0

Substiute the value of x into Equation 2 to solve for y:

y = -5x + 2

y = -5(0) + 2

y = 2

<h3><u>Solution of the given systems of linear equations:</u></h3>

Therefore, the solution to the given systems of linear equations is: x = 0, y = 2 or (0, 2).

<h2>Question 6:</h2>

\displaystyle\mathsf{\left \{\quad\:Equation\:1:\:-2x+6y=\:-30} \atop{Equation\:2:\:y-2x=\:-10}} \right.}

Add 2x to both sides of Equation 2 to isolate y:

y – 2x = –10

y – 2x + 2x = 2x – 10

y = 2x – 10

Substitute the value of y from the previous step into Equation 1:

– 2x + 6y =  – 30

– 2x + 6(2x – 10) =  – 30

Distribute 6 into the parenthesis:

– 2x + 12x  – 60 =  – 30

Combine like terms:

10x – 60 =  – 30

Add 60 to both sides:

10x – 60 + 60 =  – 30 + 60

10x = 30

\displaystyle\mathsf{\frac{10x}{10}\:=\:\frac{30}{10} }

x = 3

Substitute the value of x into Equation 2 to solve for y:

y – 2x = –10

y – 2(3) = –10

y – 6 = –10

y – 6 + 6 = –10 + 6

y = – 4

<h3><u>Solution of the given systems of linear equations:</u></h3>

Therefore, the solution to the given systems of linear equations is: x = 3, y = – 4 or (3, – 4).

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