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yarga [219]
3 years ago
8

4x-5y=39 y=6x-13how would you solve it and what system of equation is it

Mathematics
1 answer:
yawa3891 [41]3 years ago
6 0
Hi there,

4x - 5y = 39
y = 6x - 13

4x - 5(6x - 13) = 39
4x - 30x + 65 = 39
-26x = 39-65
-26x = -26
x= -26/-26
x= 1

So we deduce y now :

y = 6 *1 - 13
y= 6 - 13 = -7

Hope this helps !

Photon
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23.64112 > 23.6404

Reason : 1/1000 place value in 23.64112  is 1.

1/1000 place value in 23.6404  is 0.

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Option A) Is the answer. \boxed{\mathbf{\dfrac{3f^3}{g^2}}}

For this question; You are needed to expose yourselves to popular usages of radical rules. In this we distribute the squares as one-and-a-half fractions as the squares eliminate the square roots. So, as per the use of fraction conversion from roots. It becomes relatively easy to solve and finish the whole process more quicker than everyone else. More easier to remember.

Starting this with the equation editor interpreter for mathematical expressions, LaTeX. Use of different radical rules will be mentioned in between the steps.

Radical equation provided in this query.

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Divide the numbered values of 900 and 100 by cancelling the zeroes to get "9" as the final product in the next step.

\mathbf{\sqrt{\dfrac{9f^6}{g^4}}}

Imply and demonstrate the rule of radicals. In this context we will use the radical rule for fractions in which a fraction with a denominator of variable "a" representing a number or a variable, and the denominator of variable "b" representing a number or a variable are square rooted by a value of "n" where it can be a number, variable, etc. Here, the radical of "n" is distributed into the denominator as well as the numerator. Presuming the value of variable "a" and "b" to be greater than or equal to the value of zero. So, by mathematical expression it becomes:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}, \: \: a \geq 0 \: \: \: b \geq 0}}

\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{\sqrt{g^4}}}

Apply the radical exponential rule. Here, the squar rooted value of radical "n" is enclosing another variable of "a" which is raised to a power of another variable of "m", all of them can represent numbers, variables, etc. They are then converted to a fractional power, that is, they are raised to an exponent as a fractional value with variables constituting "m" and "n", for numerator and denominator places, respectively. So:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}

\mathbf{Since, \quad \sqrt{g^4} = g^{\frac{4}{2}}}

\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{g^2}}

Exhibit the radical rule for two given variables in this current step to separate the variable values into two new squares of variables "a" and "b" with a radical value of "n". Variables "a" and "b" being greater than or equal to zero.

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}, \: \: a \geq 0 \: \: \: b \geq 0}}

So, the square roots are separated into root of 9 and a root of variable of "f" raised to the value of "6".

\mathbf{\therefore \quad \dfrac{\sqrt{9} \sqrt{f^6}}{g^2}}

Just factor out the value of "3" as 3 × 3 and join them to a raised exponent as they are having are similar Base of "3", hence, powered to a value of "2".

\mathbf{\therefore \quad \dfrac{\sqrt{3^2} \sqrt{f^6}}{g^2}}

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\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^n} = a^{\frac{n}{n}} = a}}

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\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}

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Hi there! :)

Step-by-step explanation:

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Answer: \boxed{=3685}

Hope this helps!

Thanks!

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Have a great day! :)

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