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kramer
3 years ago
10

Resuelve el siguiente sistemas de ecuaciones por el método

Mathematics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

x=10 and y=12

Step-by-step explanation:

To solve this quadratic equation we will use two method

1. Elimination method

2. substitution method

first of we use elimination method

we either eliminate x or y

we will be eliminating y

83x-20y=590..........(eq1)

60x+18y=816...........(eq2)

y will be eliminated by multiplying

(eq1) by 9

(eq2) by 10

which will give

747x-180y=5310.........(eq3)

600x-180y=8160.........(eq4)

see that y has the same value that is (-180y and +180y)

so to eliminate y completely you have to add eq1 and eq2 because if you don't add them together you wont eliminate y

747x-180y=5310

+

600x+180y=8160

=

1347x+0=13470

1347x=13470

x=13470/1347

x=10

Now to find y we use substitution method ie put x=10 in any of the equation above(eq1,eq2,eq3, eq4) you will get the same answer

eq1

83x-20y=590..... where (x=10)

83(10)-20y =590

830-20y=590

like terms

-20y=590-830

-20y= -240

divide both sides with -20

y= -240/-20

y=12

OR

eq2

60x+18y=816..... where x=10

60(10)+18y=816

600+18y=816

like term

18y=816-600

18y=216

y=216/18

y= 12

or eq3 or eq4 you will still get the same answer......

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3 years ago
6-10 divide, another onee thank you!!​
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Answers:

10.) \displaystyle \pm{5}

9.) \displaystyle 1\frac{1}{2}

8.) \displaystyle \pm{1\frac{1}{2}}

7.) \displaystyle \pm{1\frac{1}{2}}

6.) \displaystyle \pm{\frac{1}{2}}

Step-by-step explanations:

10.) \displaystyle \frac{\sqrt{200}}{\sqrt{8}} \hookrightarrow \sqrt{25} \hookrightarrow \frac{\pm{10\sqrt{2}}}{\pm{2\sqrt{2}}} \\ \\ \boxed{\pm{5}}

9.) \displaystyle \frac{\sqrt[3]{135}}{\sqrt[3]{40}} \hookrightarrow \sqrt[3]{3\frac{3}{8}} \hookrightarrow \frac{3\sqrt[3]{5}}{2\sqrt[3]{5}} \\ \\ \boxed{1\frac{1}{2}}

8.) \displaystyle \frac{\sqrt[4]{162}}{\sqrt[4]{32}} \hookrightarrow \sqrt[4]{5\frac{1}{16}} \hookrightarrow \frac{\pm{3\sqrt[4]{2}}}{\pm{2\sqrt[4]{2}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}

7.) \displaystyle \frac{\sqrt{63}}{\sqrt{28}} \hookrightarrow \sqrt{2\frac{1}{4}} \hookrightarrow \frac{\pm{3\sqrt{7}}}{\pm{2\sqrt{7}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}

6.) \displaystyle \frac{\sqrt{12}}{\sqrt{48}} \hookrightarrow \sqrt{\frac{1}{4}} \hookrightarrow \frac{\pm{2\sqrt{3}}}{\pm{4\sqrt{3}}} \\ \\ \boxed{\pm{\frac{1}{2}}}

I am joyous to assist you at any time.

5 0
2 years ago
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