Question

X+y=7
x-y=5
solution with subsitution method

Answer 1

Topic : Linear equations in two variables

`\:`

Given Equations,

• x + y = 7 ----- (i)

• x - y = 5 ----- (ii)

`\:`

Solution,

First of all, let us take the first equation,

`\\ \longrightarrow \qquad{ {{ \sf{ x + y} = 7 \: \:}}} \\ \\ \: \:`

Subracting y from both sides we get :

`\\ \longrightarrow \qquad{ {{ \sf{ x + y - y} = 7 - y \: \:}}} \\ \: \:`

`\longrightarrow \qquad{ {{ \sf{ x } = 7 - y \: \:}}} \\ \\ \: \:`

Now, Substituting the value of x in Equation (ii) :

`\\ \longrightarrow \qquad{ {{ \sf{ x-y=5 \: \:}}}} \\ \: \:`

`\longrightarrow \qquad{ {{ \sf{ (7 - y)-y=5 \: \:}}}} \\ \: \:`

`\longrightarrow \qquad{ {{ \sf{ 7 - 2y=5 \: \:}}}} \\ \\ \: \:`

Now, Subtracting 7 from both sides :

`\\ \longrightarrow \qquad{ {{ \sf{ 7 - 2y - 7=5 - 7 \: \:}}}} \\ \: \:`

`\longrightarrow \qquad{ {{ \sf{ - 2y = - 2\: \:}}}} \\ \\ \: \:`

Dividing both sides by -2 we get :

`\\ \longrightarrow \qquad{ {{ \sf{ ( - 2y)/( - 2) = ( - 2)/( - 2) \: \:}}}} \\ \: \:`

`\longrightarrow \qquad{ \underline{\boxed{ \pmb{ \mathfrak{y = \: \: 1}}}}} \: \bigstar\\ \: </p><p>`

• Therefore, The value of y is 1

Now, Substituting the value of y in Equation (i) :

`\\ \longrightarrow \qquad{ {{ \sf{ x+y=7 \: \:}}}} \\ \: \:`

`\longrightarrow \qquad{ {{ \sf{ x+1=7 \: \:}}}} \\ \\ \: \:`

Subracting 1 from both sides we get :

`\\ \longrightarrow \qquad{ {{ \sf{ x+1 - 1=7 - 1\: \:}}}} \\ \: \:`

`\longrightarrow \qquad{ \underline {\boxed{ \pmb{ \mathfrak{x = \: \: 6}}}}} \: \: \bigstar\\ \:`

• Therefore, The value of x is 16

Answer 2

Hey ! there

Answer:

Solution of equation or we can say that value of

• x = 6

• y = 1

Explanation:

In this question we are given with two equations that are ,

• two equations that are ,x + y = 7

• two equations that are ,x + y = 7x - y = 5

And we are asked to find the solution of equation with the help of substitution method .

SOLUTION : -

Firstly we are giving numbering to the equation so that there's ease in solving . So ,

• x + y = 7 ----------- ( Equation 1 )

• x - y = 5 ----------- ( Equation 2 )

We can see that Equation 2 is smaller than Equation 1 . So we are using Equation 2 to find the value of x .

Finding value of x from Equation 2 :

• 
  `\rm{x - y = 5}`

Adding y on both sides :

• 
  `\rm{x - \cancel{y }+ \cancel{ y }= 5 + y}`

We get :

• 
  `\underline{ \boxed{ \rm{x = 5 + y}}} - - - - (\rm{Equation \: 3})`

Therefore , value of x is 5 + y .

Now substituting value of x as 5 + y in Equation 1 in order to find the value of y. So ,

`\: \quad \: \longmapsto \: \qquad \: \rm{ x + y = 7}`

Step 1 : Substituting value of x :

`\: \quad \: \longmapsto \: \qquad \: \rm{ \bold{5 + y} + y = 7}`

Step 2 : Adding like terms that are y and y :

`\: \quad \: \longmapsto \: \qquad \: \rm{5 + 2y = 7}`

Step 3 : Subtracting 5 on both sides :

`\: \quad \: \longmapsto \: \qquad \: \rm{ \cancel{5} + 2y - \cancel{5} = 7 - 5}`

We get ,

`\: \quad \: \longmapsto \: \qquad \: \rm{2y = 2}`

Step 4 : Dividing both sides with 2 :

`\: \quad \: \longmapsto \: \qquad \: \rm{ \frac{ \cancel{2}y}{ \cancel{2}} = \cancel{ (2)/(2) }}`

On simplifying, We get :

`\: \quad \: \longmapsto \: \qquad \: \blue{ \underline{\boxed{ \frak{y = 1}}}} \quad \bigstar`

• Henceforth , value of y is ❝ 2 ❞

Now finding value of x from Equation 3 :

For finding value of x we are substituting value of y in Equation 3 . So ,

`\: \quad \: \longmapsto \: \qquad \: \rm{x = 5 + y}`

Substituting value of y :

`\: \quad \: \longmapsto \: \qquad \: \rm{x = 5 + 1}`

Adding 5 with 1 , We get :

`\: \quad \: \longmapsto \: \qquad \: \blue{\underline{\boxed{\frak{x = 6}}}} \quad \bigstar`

• Henceforth , value of x is ❝ 6 ❞

From values of x as 6 and y as 1 we can say that they are the solution of given equations .

Verifying : -

Now we are checking our answer whether it is wrong or right .

Equation 1 : x + y = 7

Substituting value of x and y in Equation 1 :

• 6 + 1 = 7

• 7 = 7

• L.H.S = R.H.S

• Hence, Verified.

Therefore, our answer is correct .

Equation 2 : x - y = 1

• 6 - 1 = 5

• 5 = 5

• L.H.S = R.H.S

• Hence , Verified .

Therefore , our answer is correct .

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