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ira [324]
2 years ago
6

What is the best first step for solving the given system using substitution while avoiding fractions?

Mathematics
2 answers:
lapo4ka [179]2 years ago
6 0
Well, it says to avoid fractions. So I did the math: 

Solving for x in the first equation get you: x= \frac{7}{2}-3y

Solving for y in the first equation gets you:y= \frac{7}{6} - \frac{2}{6}x

Solving for x in the second equation gets you: x=4-3y

Solving for y in the second equation gets you:y= \frac{24}{18} - \frac{6}{18}x

If you look you can see the only option that does not give you fractions is if you solve for x in the second equation. So your answer is c! I hope this helps! =) 
sergey [27]2 years ago
6 0
Solve with subsitution avoiding fractions

<span>2x+6y=7
6x+18y=24

if we solve for x in first, we get x=-3y+7/2, has fractions so no
if we solve for y in first, we get y=-x/3+7/6, has fractions so no
if we solve for x in the 2nd, we get x=-3y+4, no fractions so yay
if we solve for y in the 2nd, we get y=-x/3+4/3, has fractions so no


answer is C
</span>
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Please help me..........<br><br>only 32,and 33​
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Answer:

Step-by-step explanation:

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Step 1 : Draw PQ = 6cm

Step 2 : Construct 60 degree at Q. [Given PQR = 60=> Q = 60 ]

Step 3 : Take 5 cm on compass and mark 5cm on the 60 degree

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Step 4 : The arc of 5cm and 60 degree line  meets is the point R.

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Step 5 : Given PQ || SR so construct a 120° at R ,

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33

Given adjacent sides : 4.8 and 4.2

And it is a rectangle.

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8 0
3 years ago
Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?
worty [1.4K]

Answer:

4

Step-by-step explanation:

(x+y+z)(xy+xz+yz)=18 Equation 1

x^2(y+z)+y^2(x+z)+z^2(x+y)=6 Equation 2

What is the value of xyz where each variable represents a real number?

Let's expand equation 1:

(x+y+z)(xy+xz+yz)=18

x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18

Simplify each term if can:

x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have x^2 and if I factored x^2 from first two terms I would have x^2(y+z) which is the first term of left hand side of equation 2.

So let's see what happens if we gather the terms together that have the same variable squared together.

x^2y+x^2z+y^2x+y^2z+z^2y+z^2x+xyz+xyz+xyz=18

Factor the variable squared terms out of each binomial pairing:

x^2(y+z)+y^2(x+z)+z^2(y+x)+xyz+xyz+xyz=18

Replace the sum of those first three terms with what it equals which is 6 from the equation 2:

6+xyz+xyz+xyz=18

Combine like terms:

6+3xyz=18

Subtract 6 on both sides:

3xyz=12

Divide both sides by 3:

xyz=4

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Bezzdna [24]

Answer:

3.5 is the correct answer

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