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3 years ago

How do you solve x-y=1 ; x+y=3. Please Explain

2 answers:

8 0

You isolate one variable. Let's say that x will be that variable in the first equation:
x-y=1
+y +y
x=y+1
So now you have the equation for x. All you have to do is substitute it into the other:
x=y+1
(y+1)+y=3
Combine like terms and solve:
2y+1=3
-1 -1
2y=2
2y/2=2/2
y=1
Now that you have the value of one, substitute it into either of the equations to find x.
If you put it in the first:
x-1=1
+1 +1
x=2
If you put it into the other, you would get the same answer.
Therefore, y=1 and x=2

zepelin [54]3 years ago

4 0

X - y = 1
x + y = 3
________
2x = 4
__ _
2 2
x = 2
x + y = 3
2 + y = 3
-2 -2
________
y = 1
(x, y) = (2, 1)

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valentina_108 [34]

Answer:

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Step-by-step explanation:

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2 years ago

Find the arc length, x, when<br> r= 8 and 0 = 5 radians.

lys-0071 [83]

Answer:

Arc Length = 40

Step-by-step explanation:

Arc Length Formula: S = rΘ

R = radius

Θ = Radians

<u>Step 1: Identify the radius and the radian angle</u>

R = 8

Θ = 5 radians

<u>Step 2: Plug into the equation</u>

S = (8)(5)

<u>Step 3: Solve</u>

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3 years ago

⭐Whoever that Answers all of These First, Gets Brainliest !!! (Easy Edition)⭐

serious [3.7K]

Answer:

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Step-by-step explanation:

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3 years ago

when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.

sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

<h3><u>Solution:</u></h3>

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = $a^2$

6 is subtracted from the square of a number = $a^2 - 6$

5 times the number = $5 \times a$

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

$a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0$

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here in this problem,

$a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6$

Substituting the values in above quadratic formula, we get

$\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}$

We have two solutions for "a"

$\begin{array}{l}{a=\frac{5+7}{2} \text { and } a=\frac{5-7}{2}} \\\\ {a=\frac{12}{2} \text { and } a=\frac{-2}{2}}\end{array}$

We have asked negative solution. So a = -1

Thus the negative solution is -1

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3 years ago

Please help!! <br><br> The length of MO is____ units.

aivan3 [116]

Use the Pyth. Thm. to calculate the length of MO:

6^2 + 8^2 = |MO| = 36 + 64 = 100. Therefore, |MO| = 10.

4 0

3 years ago

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