Hey friend, hope I can assist you!
I will solve by elimination <3.
Multiply 3x - 7y = 2 by 2: 6x - 14y = 4
6x - 14y = 4
6x - 9y = 9
5y = 5
Now we have
6x - 14y = 4
5y = 5
Now we want to solve 5y = 5 for y
So simply divide both sides by 5.
5y/5 = 5/5
This gives us one or in other words, y = 1.
Now we want to plug y = 1 into 6x - 14y = 4
So 6x - 14 * 1 = 4
This gives us
6x - 14 = 4
Now add 14 to both sides.
6x - 14 + 14 = 4 + 14
6x = 18
Now divide both sides by 6
6x/6 = 18/6
This gives us 3 so x = 3
Therefore our solutions to this system of equations would be y = 1 and x = 3
Answer:
(2x-3)(2x+1)
Step-by-step explanation:
4x^2 - 4x - 3
we need two numbers that give 12 when multiplied and 4 when subtracted
the numbers are 6 and 2 because 6*2=12 and 6-4=4.
4x^2 - (6-2)x -3
4x^2 -6x +2x -3
take common
2x(2x -3)+1(2x-3)
take (2x-3) as common
(2x-3)(2x*1 + 1*1)
(2x-3)(2x+1)
The quotient of the number given number 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.
<h3>What is the quotient?</h3>
Quotient is the resultant number which is obtain by dividing a number with another. Let a number a is divided by number b. Then the quotient of these two number will be,

Here, (<em>a, b</em>) are the real numbers.
The number StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction, given can be written as,

Let the quotient of this division is n. Therefore,

A number in numerator of a fraction with negative exponent can be written in the denominator with the same but positive exponent and vise versa. Therefore,

Hence, the quotient of the number given number 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.
Learn more about the quotient here;
brainly.com/question/673545
If n is rational, it means that

Therefore when we do n² we can write it as

Remember that the product of two integer numbers is also an integer, therefore we can guarantee that

Then we can confirm that n² is the quotient of two integers and the denominator is not zero, therefore, n² is always rational, it cannot be an irrational number