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3 years ago

2x-y=4 and y=-2x+8. Can anyone help me solve

Mathematics

2 answers:

Komok [63]3 years ago

5 0

Answer:

The correct answer is x = 3 and y = 2.

Step-by-step explanation:

There are many ways to solve systems of equations like this, but I'm going to use substitution. This means taking the value of y given by the second equation and plugging it into the first equation. This is modeled below:

2x - y = 4

2x - (-2x+8) = 4

Now, we can simplify the left side of the equation.

2x + 2x - 8 = 4

4x - 8 = 4

We should add 8 to both sides as the next step.

4x = 12

Now we can divide by 4.

x = 3

To solve for y, we can substitute this value found for x back into either one of our original equations.

y = -2x + 8

y = (-2*3) + 8

y = -6 + 8

y = 2

Therefore, the correct answer is x = 3 and y = 2.

Hope this helps!

musickatia [10]3 years ago

3 0

Answer:

The value of x is 3 and y is 2

Step-by-step explanation:

Here, 2x - y = 4 and y = -2x + 8 are two given equation,

Now, We will use Substitution Method

2x - y = 4 ...(1)

y = -2x + 8 ...(2)

Substitute the value of y = -2x + 8 in equation (1)

2x - y = 4

2x - (-2x + 8) = 4

2x + 2x - 8 = 4

4x - 8 = 4

Add 8 both side we get,

4x - 8 + 8 = 4 + 8

4x = 12

Divide by 4 both side we get,

4x/4 = 12/4

x = 3

Now, Substitute the value of x in equation (2)

y = -2x + 8

y = -2(3) + 8

y -6 + 8

y = 2

Thus, The value of x is 3 and y is 2

<h3>FOR VERIFICATION ONLY:</h3>

2x - y = 4

2(3) - (2) = 4

6 - 2 = 4

4 = 4

y = -2x + 8

2 = -2(3) + 8

2 = -6 + 8

2 = 2

Hence, L.H.S = R.H.S

-TheUnknownScientist

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MArishka [77]

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3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

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$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

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$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

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$c$ is another arbitrary prime number. (3)

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