Question

Three times of a number added to two times of another number gives 33. When three times of the second number is subtracted from four times the other number got 10. taking the numbers as x and y form the equations and find the numbers?

Answer 1

Given :

• Three times of a number added to two times of another number gives 33.
• Three times of second number subtracted from four times the first number gives 10.

To find :

• Two numbers

Solution

Let the numbers be x and y

A/q to 1st condition,

⇒ 3x + 2y = 33                           __(i)

A/q to 2nd condition,

⇒ 4x - 3y = 10                            __(ii)

Now multiplying (i) into 3 and (ii) into 2, we get,

                         9x + 6y = 99

                         8x - 6y  = 20

                      (+)__(-)___(+)_

                          17x = 119

                        ⇒ x = 119/17

                        ⇒ x = 7

Now putting value of x in (i) :

⇒ 3(7) + 2y = 33

⇒ 21 + 2y = 33

⇒ 2y = 33 - 21

⇒ 2y = 12

⇒ y = 12/2

⇒ y = 6

Therefore,

First number is 7 and second number is 6 .

Answer 2

Answer :-

$\: \: \mapsto \: \: \underline{\boxed{\sf{\red{Let's \: understand \: the \: concept}}}}$

Here the concept of Linear Equations in Two Variables has been used. According to this, if we make the value of one variable depend on other from any equation, we can easily get the values of both. The standard form of Linear Equations in Two Variable is given as :-

ax + by + c = 0

px + qy + c = 0

Using this concept, let us do this question.

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★ Question :-

Three times of a number added to two times of another number gives 33. When three times of the second number is subtracted from four times the other number got 10. Taking the numbers as x and y form the equations and find the numbers ?

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★ Solution :-

Given,

» 3 × the first number + 2 × another number = 33

» 4 × first number - 3 × the second number = 10

• Let the first number be 'x' and the second number be 'y', then

According to the given things and variables,

~ Case I :-

➮ 3x + 2y = 33

➮ 2y = 33 - 3x

➮ y = ½(33 - 3x) ...(i)

~ Case II :-

➮ 4x - 3y = 10 ... (ii)

From equation (i) and equation (ii), we get,

➮ 4x - 3 × ½(33 - 3x) = 10

Now let us multiply each and every term by 2, then,

➮ 2 × 4x - 2 × 3 × ½(33 - 3x) = 2 × 10

➮ 8x + 9x - 99 = 20

➮ 17x = 20 + 99

➮ 17x = 119

$\: \: \huge{\Longrightarrow \: \: \bold{x} \: = \: \dfrac{119}{17}}$

➮ x = 7

Hence, the first number is = x = 7

From equation (i) and value of x, we get,

➠ y = ½(33 - 3x)

➠ y = ½(33 - 3(7))

➠ y = ½(33 - 21)

➠ y = ½(12)

➠ y = 6

Hence, the other number is = y = 6

From this, we get,

$\: \: \: \: \underline{\boxed{\rm{\blue{\mapsto \: \: \: Thus \: the \: first \: number \: is \: \underline{7} \: and \: the \: second \: number \: is \: \underline{6}}}}}$

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$\: \: \underline{\boxed{\sf{\green{Let's \: verify \: it}}}}$

For verification, we need to simply apply the values we got, into the equations we formed. So,

~ Case I :-

=> 2y = 33 - 3x

=> 2(6) = 33 - 3(7)

=> 12 = 33 - 21

=> 12 = 12

Clearly, LHS = RHS.

~ Case II :-

=> 4x - 3y = 10

=> 4(7) - 3(6) = 10

=> 28 - 18 = 10

=> 10 = 10

Clearly, LHS = RHS

Here both the conditions satisy, so our answer is correct.

Hence, verified.

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$\: \: \underline{\boxed{\rm{Let's \: understand \: more}}}$

• Linear Equations are the equations formed using constant and variable terms but of single degree that is 1.

• Polynomial is also a equation formed using variable term but can be of many degrees.

• Linear Equation in Two Variable are the equation formed using both constant and variable term but has single degree and two variables.