The sum of two positive integers is 31. If the sum of the squares of these numbers is 625,find the smaller of the numbers.

Answers

Answer 1

Step-by-step explanation:

It is given that, the sum of two positive integers is 31 and the sum of the squares of these numbers is 625 and we are to find the smaller of the numbers.

So, let the two positive integers be x and y.

Therefore,

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 \: \: ........ \: (i) }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + y {}^{2} = 625 \: ...... \: (ii)}}} \\ \\[/tex]

Now, From the first equation we have,

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 }}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - x \: ...... \: (iii)}}} \\ \\ [/tex]

Now, substituting the value of y in equation (ii) we get :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 - x) {}^{2} = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 {}^{2} - 2.x.31 + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (961 - 62x + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + 961 - 62x + x{}^{2} - 625 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} + 336 - 62x = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} - 62x + 336 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2( x {}^{2} - 31 + 168) = 0}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad \frac{2}{2} ( x {}^{2} - 31 + 168) = \frac{0}{2} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} - 31 + 168 = 0}}} \\ \\[/tex]

Now using the quadratic formula :

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}} \\ \\[/tex]

Where,

a = 1

b = -31

c = 168

[tex] \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - (- 31) \pm \sqrt{ {31}^{2} - 4(1)(168)} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 961 -672} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 289} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm 17 }{2(1)} }}} \\ \\[/tex]

Now, we have two equations,

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} \: ... .....\: (iv)}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2} ... .....\: (v)}}} \\ \\[/tex]

So, Equation (iv) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 48 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = 24 }}} \\ \\ [/tex]

Now, Equation (v) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 14 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \: 7}}}\\ \\ [/tex]

So, the value of x is 7 or 24

Now, we are to find the value of y.

Substituting the value of x (24) in equation (iii) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 31 - 24 \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 7 \:}}} \\ \\ [/tex]

Again, Substituting the value of x (7) in equation (iii) :

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - 7 \:}}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 24 \:}}} \\ \\ [/tex]

Therefore,

The value of y is also 7 or 24.

[tex] \\ [/tex]

So, The smaller of the numbers is 7 .

Answer 2
Question : -

The sum of two positive integers is 31. If the sum of the squares of these numbers is 625, find the smaller of the numbers.

Given : -

Sum of two positive numbers = 31

Sum of squares of these numbers = 625

To Find : -

We have to find the smaller of the numbers .

Concept : -

This question belongs to quadratic equations so we have to find the answer by making equation and solving it .

To Assume : -

Let the first no. be x

Let the second no. be y

So let's get started with Solution :

According to question , sum of two positive integers is 31 . So ,

x + y = 31 --------- ( Equation 1 )

According to question , sum of square

of these numbers is 625 . So ,

x² + y² = 625 --------- ( Equation 2 )

From equation 1 ( x + y = 31 ) , Value

of x :

x = 31 - y

Now , putting value of x in eq. 2 :

x² + ( 31 - x )² = 625

x² + ( 31 )² - ( 2 × 31 × x ) + x² = 625

2x² + 961 - 62x = 625

2x² - 62x = 625 - 961

2x² - 62x = -336

2x² - 62x + 336 = 0

2(x² - 31x + 168 ) = 0

x² - 31x + 168 = 0

Solving it by using middle term

splitting :

x² -24x -7x + 168 = 0

x ( x - 24 ) -7 ( x - 24 ) = 0

( x - 7 ) ( x - 24 )

So ,

First number ,

x - 7 = 0x = 7 { Smaller Number }

Second Number ,

x - 24 = 0x = 24

Verification :

According to question ,

Sum of numbers is equal to 31 :

x + y = 31

7 + 24 = 31

L.H.S = R.H.S

Sum of squares of these numbers is equal to 625 :

x² + y² = 625

7² + 24² = 625

49 + 576 = 625

625 = 625

L.H.S = R.H.S

Therefore , our value for x and y are true. Thus our answer is valid.

#[tex] \rm{Keep \: Learning}[/tex]

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Step-by-step explanation:

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Step-by-step explanation: see attached.




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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

[tex]136.96 \div 12.8 [/tex]

that show give you your answer

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Answers

Step-by-step explanation:

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MARK ME AS BRAINLIST

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Answers

hello!

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[tex]\sf{-3x < 12}[/tex]

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[tex]\bigstar{\boxed{\pmb{The~numbers~greater~than~-4}}}[/tex]

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note:-

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Answers

Answer:

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Step-by-step explanation:

Given:

[tex]\displaystyle \large{-\dfrac{1}{125}}[/tex]

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Answers

The length of fence materials needed to cover a land of sides 700m ,700m and 850m is 2250 meters

Perimeter of Flat Shapes

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Answers

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Below!

Step-by-step explanation:

Let's simplify each side to determine if both sides are equal.

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Answers

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Answers

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Step-by-step explanation:

REFER TO THE ATTACHMENT!!

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Answer:

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Answers

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Step-by-step explanation:

rise/run or up and down/left and right

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Write an equation to determine how many ounces of cheese, ccc, Noah can use on each sandwich.

Answers

Answer:

24 / 6 = ccc

Step-by-step explanation:

There is 24 oz of cheese total that will be split evenly (divided) into 6 parts.

Each part will equal one sandwich worth of cheese (ccc) on the other side of the equation.

This one is very simple to solve, you could just calculate 24/6 to get your answer. But I'll go through it using typical algebra you would use for a less simple equation.

To solve:

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Answer:

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Answer:

14π inches

Step-by-step explanation:

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527.52 in

Step-by-step explanation:

To answer this question you have to find the circumference of the circle.

You have to use the formula below to find the circumference.

C = π d

Here,

d ⇒ diameter ⇒ 7 in

Let us solve it now.

C = π × d

C = 3.14 × 7

C =  21.98 in

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Answers

Answer:

136.

Step-by-step explanation:

-8 . (-10+(-7))

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= 80 + 56

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