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MissTica
2 years ago
14

The sum of numbers is 28. the difference of two numbers is -2. find the numbers​

Mathematics
2 answers:
Vesna [10]2 years ago
7 0
I believe 15 and 13. 13-15=-2 and added is 28....?
Phantasy [73]2 years ago
4 0

Answer:

x=13 and y=15

Step-by-step explanation:

Let the 2 numbers be x and y

ATQ,

x+y = 28 ------ eqn.1

x-y=-2 -------- eqn.2

Adding both the equations:

x+y=28

+x-y=-2

_______

2x = 26

Therefore x=13

Now substituting the value of x in eqn. 1

x+y=28

13+y=28

Therefore y=28-13

=15

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It is False because t is multiplied by a negative number, so to solve for it you will need to divide both sides by -3 which mean you have to flip the sign. 

Hope this helps!
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If f(x, y)=xy−x^2, what is the value of f(f(−2, 3), −4)?
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Is there any app which helps solve pure maths questions?​
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2 years ago
PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%
Yanka [14]

Answer:

x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}

Second equation:

\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}

Third equation:

\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}

Now we use this value for "x" back in equation 1 to solve for "y":

1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}

And finally we solve for the third unknown "z":

\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12

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2 years ago
120 books increased by 55%
insens350 [35]
The answer would be 186
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2 years ago
Read 2 more answers
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