Question

If three times the sum of two numbers is 42and five time's their different is 20,find the number.

Answer 1

Given :

• Three times the sum of two numbers is 42

• Five times their difference is 20

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‎

To find :

• The two numbers

‎

‎

Concept :

In this question we are given two criteria and we are asked to find the two numbers which would satisfy the given criterias . For this we would frame up two equation following both the criterias . Then solving those linear simultaneous equation by Elimination method we would get out final answers .

Elimination method :

Its a type of method used to solve linear simultaneous equation . Steps included are -

Firstly the leading coefficient should be same if not we need to multiply by certain numbers to make it same.

Secondly we need to subtract equation (ii) from equation (i) . Doing so we would get the value of one variable.

Finally substituting the value of the variable in other equation we will get the other variable also.

Hope am clear let's solve :D~

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Assumption :

Leet the two numbers be x and y .

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Solution :

Forming equations -

Criteria 1

Three times the sum of two numbers is 42

$\tt\:3 \times (x+y) =42$

Transposing 3 to RHS it goes to the denominator

$\tt\to\:(x+y) =\frac{42}{3}$

Reducing the fraction to lower terms

$\tt\to\:(x+y) =\cancel{\frac{42}{3}}$

$\tt\to\:x+y=14$ $\sf\color{blue}{--(i)}$

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Criteria 2

Five times the difference of two numbers is 20

$\tt\:5 \times (x-y) =20$

Transposing 5 to RHS it goes to the denominator

$\tt\to\:(x-y) =\frac{20}{5}$

Reducing the fraction to lower terms

$\tt\to\:(x-y) =\cancel{\frac{20}{5}}$

$\tt\to\:x-y=4$ $\sf\color{blue}{--(ii)}$

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As we see in both the equation leading term is same i.e., x. So we need to multiply. Directly jumping into second step i.e., subtraction.

Subtracting equation (ii) from equation (i)

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$\tt\:Equation~(i)~-~Equation~(ii)$

$\tt\implies\:x+y-(x-y)=14-4$

Multiplying the signs and removing the brackets

$\tt\implies\:x+y-x+y=14-4$

Arranging the like terms together

$\tt\implies\:x-x+y+y=14-4$

Terms with opposite signs gets cancelled

$\tt\implies\:\cancel{x}-\cancel{x}+y+y=14-4$

$\tt\implies\:y+y=14-4$

$\tt\implies\:2y=10$

Transposing 2 to RHS it goes to the denominator

$\tt\to\:y =\frac{10}{2}$

Reducing the fraction to lower terms

$\tt\to\:y =\cancel{\frac{10}{2}}$

$\small{\underline{\boxed{\mathrm{\implies\:y=5}}}}$ ★

‎

Substituting the value of y as 5 in equation (i) -

$\tt\:x+y=14$ $\sf\color{blue}{--(i)}$

$\tt\implies\:x+5=14$

Transposing +5 to RHS it becomes -5

$\tt\implies\:x=14-5$

$\small{\underline{\boxed{\mathrm{\implies\:x=9}}}}$ ★

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Verifying :

Plugging both the values of x as 9 and y as 5 in equation (i) -

$\tt\:x+y=14$

$\tt\leadsto\:9+5=14$

$\tt\leadsto\:14=14$

$\bf\leadsto\:LHS=RHS$

Hence Verified !~

______________________

Henceforth,

‎ ⠀⠀ ⠀⠀ ⠀⠀❒ Value of x = $\large{\mathfrak\red{9}}$

‎ ⠀⠀ ⠀⠀ ⠀⠀❒ Value of y = $\large{\mathfrak\red{5}}$

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$\dag\:\underline{\sf So\:the\:two\:numbers\:are\:9\:and\:5}$

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Answer 2

Step-by-step explanation:

Given :

Three times the sum of two numbers is 42

Five times their difference is 20

‎

‎

To find :

The two numbers

‎

‎

Concept :

In this question we are given two criteria and we are asked to find the two numbers which would satisfy the given criterias . For this we would frame up two equation following both the criterias . Then solving those linear simultaneous equation by Elimination method we would get out final answers .

Elimination method :

Its a type of method used to solve linear simultaneous equation . Steps included are -

Firstly the leading coefficient should be same if not we need to multiply by certain numbers to make it same.

Secondly we need to subtract equation (ii) from equation (i) . Doing so we would get the value of one variable.

Finally substituting the value of the variable in other equation we will get the other variable also.

Hope am clear let's solve :D~

‎

Assumption :

Leet the two numbers be x and y .

‎

Solution :

Forming equations -

Criteria 1

Three times the sum of two numbers is 42

\tt\:3 \times (x+y) =423×(x+y)=42

Transposing 3 to RHS it goes to the denominator

\tt\to\:(x+y) =\frac{42}{3}→(x+y)=

3

42

Reducing the fraction to lower terms

\tt\to\:(x+y) =\cancel{\frac{42}{3}}→(x+y)=

3

42

\tt\to\:x+y=14→x+y=14 \sf\color{blue}{--(i)}−−(i)

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Criteria 2

Five times the difference of two numbers is 20

\tt\:5 \times (x-y) =205×(x−y)=20

Transposing 5 to RHS it goes to the denominator

\tt\to\:(x-y) =\frac{20}{5}→(x−y)=

5

20

Reducing the fraction to lower terms

\tt\to\:(x-y) =\cancel{\frac{20}{5}}→(x−y)=

5

20

\tt\to\:x-y=4→x−y=4 \sf\color{blue}{--(ii)}−−(ii)

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As we see in both the equation leading term is same i.e., x. So we need to multiply. Directly jumping into second step i.e., subtraction.

Subtracting equation (ii) from equation (i)

‎

\tt\:Equation~(i)~-~Equation~(ii)Equation (i) − Equation (ii)

\tt\implies\:x+y-(x-y)=14-4⟹x+y−(x−y)=14−4

Multiplying the signs and removing the brackets

\tt\implies\:x+y-x+y=14-4⟹x+y−x+y=14−4

Arranging the like terms together

\tt\implies\:x-x+y+y=14-4⟹x−x+y+y=14−4

Terms with opposite signs gets cancelled

\tt\implies\:\cancel{x}-\cancel{x}+y+y=14-4⟹

x

−

x

+y+y=14−4

\tt\implies\:y+y=14-4⟹y+y=14−4

\tt\implies\:2y=10⟹2y=10

Transposing 2 to RHS it goes to the denominator

\tt\to\:y =\frac{10}{2}→y=

2

10

Reducing the fraction to lower terms

\tt\to\:y =\cancel{\frac{10}{2}}→y=

2

10

\small{\underline{\boxed{\mathrm{\implies\:y=5}}}}

⟹y=5

★

‎

Substituting the value of y as 5 in equation (i) -

\tt\:x+y=14x+y=14 \sf\color{blue}{--(i)}−−(i)

\tt\implies\:x+5=14⟹x+5=14

Transposing +5 to RHS it becomes -5

\tt\implies\:x=14-5⟹x=14−5

\small{\underline{\boxed{\mathrm{\implies\:x=9}}}}

⟹x=9

★

‎

Verifying :

Plugging both the values of x as 9 and y as 5 in equation (i) -

\tt\:x+y=14x+y=14

\tt\leadsto\:9+5=14⇝9+5=14

\tt\leadsto\:14=14⇝14=14

\bf\leadsto\:LHS=RHS⇝LHS=RHS

Hence Verified !~

______________________

Henceforth,

‎ ⠀⠀ ⠀⠀ ⠀⠀❒ Value of x = \large{\mathfrak\red{9}}9

‎ ⠀⠀ ⠀⠀ ⠀⠀❒ Value of y = \large{\mathfrak\red{5}}5

‎

‎

\dag\:\underline{\sf So\:the\:two\:numbers\:are\:9\:and\:5}†

Sothetwonumbersare9and5

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