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tatiyna
3 years ago
13

Solve the simultaneous equations: 3x + 2y = 35 2x + 3y = 30

Mathematics
1 answer:
lesya [120]3 years ago
5 0

Answer:

nuber 1

Simplifying

3x + 2y = 35

Solving

3x + 2y = 35

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-2y' to each side of the equation.

3x + 2y + -2y = 35 + -2y

Combine like terms: 2y + -2y = 0

3x + 0 = 35 + -2y

3x = 35 + -2y

Divide each side by '3'.

x = 11.66666667 + -0.6666666667y

Simplifying

x = 11.66666667 + -0.6666666667y

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alex41 [277]

So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

x-2=\sqrt{x-2}

Next, square both sides:

(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2

Next, subtract x and add 2 to both sides of the equation:

x^2-5x+6=0

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

x^2-2x-3x+6=0

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

x(x-2)-3(x-2)=0

Now you can rewrite the equation as (x-3)(x-2)=0

Now, apply the Zero Product Property and solve for x as such:

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Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}

Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>

8 0
3 years ago
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Step-by-step explanation:

You'll notice that the higher the numbers get in the experiment the closer it gets to your solution. The theoretical probability of flipping a coin is about 50% heads and 50% tails, but it doesn't always seem like that in an experiment. The experimental probability from your experimentation so far would be 62% of heads and 38% of tails.

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Suppose xy=−4 and dy/dt=−3. Find dx/dt when x=−1.
user100 [1]
When x=-1: \quad (-1)y=-4\qquad\to\qquad y=4\)

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

x'y+xy'=0

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.

From here, plug in all the stuff you know:
y' = -3
x = -1
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and solve for x'.

Hope that helps!
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vova2212 [387]

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Answer:

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