Sum points lol. I'm being generous :)

Please help me with this. Use the combination method of substitution to solve. Please show and explain work. Thanks.

Artemon [7]

Answer:

x = 25.35 (or 2129/84) and y = 4334.04 (or 121353/28)

Step-by-step explanation:

The given equations are set up and ready to go with substitution. Simply just plug in the first equation to the second equation as both are equal to y.

Step 1: Replace y in y = 87x + 2129 with 171x

171x = 87x + 2129

Step 2: Subtract 87 x on both sides

84x = 2129

Step 3: Divide both sides by 84 to get x

x = 2129/84 or 25.35 (rounded)

To get y, simply plug in x into one of the 2 original equations. In this case, I will use the first equation:

y = 171 (25.35)

y = 121353/28 or 4334.04 (rounded)

You can check your work by plugging both solutions into the calculator and see if they equal each other. The values for these answers are solely based on the equations, so if you write the equations wrong themselves, then that means you have the values wrong as well.

6 0

3 years ago

Why is cube root of 9 equal to 9 to the one third power question mark

kozerog [31]

Answer:

3

just because

3 0

2 years ago

Read 2 more answers

A-(buc)=(A-b)u(A-c)

kupik [55]

Step-by-step explanation:

I don't get the question a bit but I hope this makes sense

:)

5 0

3 years ago

Plz help out!! And no this is not a test :)

jolli1 [7]

9514 1404 393

Answer:

B = (-4, 5)

Step-by-step explanation:

Reflection over the origin changes the signs of the coordinates.

(x, y) ⇒ (-x, -y)

(4, -5) ⇒ (-4, 5)

Point B is (-4, 5).

5 0

2 years ago

Read 2 more answers

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":

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

8 0

3 years ago