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3 years ago

Which ordered pair (x, y) is a solution to given system of linear equations?

Mathematics

2 answers:

Jet001 [13]3 years ago

7 0

Answer: (1,-2)

Step-by-step explanation:

Given : the system of linear equations :-

$3x+y=1---------------(1)\\5x+y = 3---------------(2)$

In order to find x and y , first we subtract equation (1) from equation (2) , we get

$2x=2$

Divide both sides by 2 , we get

$x=1$

Substitute the value of x=1 in equation (1) , we get

$3(1)+y=1\\\\3+y=1$

Subtract 3 from both sides , we get

$y=-2$

Hence, the ordered pair (x, y) is a solution to given system of linear equations = (1,-2)

Thus , the correct answer is (1,-2) .

riadik2000 [5.3K]3 years ago

3 0

Answer:

(1, - 2 )

Step-by-step explanation:

Given the 2 equations

3x + y = 1 → (1)

5x + y = 3 → (2)

Subtracting (1) from (2) term by term eliminates the term in y, that is

(5x - 3x) + (y - y) = (3 - 1) and simplifying

2x = 2 ( divide both sides by 2 )

x = 1

Substitute x = 1 in either of the 2 equations for corresponding value of y

Using (1), then

3 + y = 1 ( subtract 3 from both sides )

y = - 2

Solution is (1, - 2 )

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3 years ago

One more than one- fourth the square of a number x is y

Tatiana [17]

Answer:

y = 1 + ¼x²

Step-by-step explanation:

Given

One more than one-fourth the square of a number x is y

Required

Represent algebraically

We start by analysing the expression one after the other.

First, the two numbers to compare are represented by x and y.

Where y is the resulting number.

So, we start with y =;

Then we move backwards

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- one-fourth "the square of a number, x".

The expression in quote has already been represented by x².

So, we can replace this expression with x².

This will give us: one-fourth of x²

One-fourth means ¼

Hence, one-fourth of x² = ¼ * x²

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In other words; 1 + ¼ * x²

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3 years ago

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Dvinal [7]

Question 11)

Answer:

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Step-by-step explanation:

Given the expression

$\log _2\left(63\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(\frac{63}{9}\right)$

$=\log _2\left(\frac{63}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{63}{9}=7$

$=\log _2\left(7\right)$

$=2.80$

Therefore,

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Question 12)

Answer:

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

Step-by-step explanation:

Given the expression

$\log _2\left(3\right)+\log _2\left(15\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _2\left(3\right)+\log _2\left(15\right)=\log _2\left(3\cdot \:15\right)$

$=\log _2\left(3\cdot \:15\right)-\log _2\left(9\right)$

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:15=45$

$=\log _2\left(45\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(45\right)-\log _2\left(9\right)=\log _2\left(\frac{45}{9}\right)$

$=\log _2\left(\frac{45}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{45}{9}=5$

$=\log _2\left(5\right)$

$=2.32$

Therefore,

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

3 0

3 years ago