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TheAmethyst

Hey ! there

Answer:

Solution of equation or we can say that value of

  • x = 6

  • y = 1

Step-by-step 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 :

  • [tex] \rm{x - y = 5}[/tex]

Adding y on both sides :

  • [tex]\rm{x - \cancel{y }+ \cancel{ y }= 5 + y} [/tex]

We get :

  • [tex] \underline{ \boxed{ \rm{x = 5 + y}}} - - - - (\rm{Equation \: 3})[/tex]

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 ,

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{ x + y = 7}[/tex]

Step 1 : Substituting value of x :

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{ \bold{5 + y} + y = 7}[/tex]

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

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{5 + 2y = 7}[/tex]

Step 3 : Subtracting 5 on both sides :

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{ \cancel{5} + 2y - \cancel{5} = 7 - 5}[/tex]

We get ,

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{2y = 2}[/tex]

Step 4 : Dividing both sides with 2 :

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{ \dfrac{ \cancel{2}y}{ \cancel{2}} = \cancel{ \dfrac{2}{2} }}[/tex]

On simplifying, We get :

[tex] \: \quad \: \longmapsto \: \qquad \: \blue{ \underline{\boxed{ \frak{y = 1}}}} \quad \bigstar[/tex]

  • 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 ,

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{x = 5 + y}[/tex]

Substituting value of y :

[tex] \: \quad \: \longmapsto \: \qquad \: \rm{x = 5 + 1}[/tex]

Adding 5 with 1 , We get :

[tex] \: \quad \: \longmapsto \: \qquad \: \blue{\underline{\boxed{\frak{x = 6}}}} \quad \bigstar[/tex]

  • 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 .

#Keep Learning

Topic : Linear equations in two variables

[tex] \: [/tex]

Given Equations,

  • x + y = 7 ----- (i)

  • x - y = 5 ----- (ii)

[tex] \: [/tex]

Solution,

First of all, let us take the first equation,

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

Subracting y from both sides we get :

[tex] \\ \longrightarrow \qquad{ {{ \sf{ x + y - y} = 7 - y \: \:}}} \\ \: \:[/tex]

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

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

[tex] \\ \longrightarrow \qquad{ {{ \sf{ x-y=5 \: \:}}}} \\ \: \:[/tex]

[tex]\longrightarrow \qquad{ {{ \sf{ (7 - y)-y=5 \: \:}}}} \\ \: \:[/tex]

[tex]\longrightarrow \qquad{ {{ \sf{ 7 - 2y=5 \: \:}}}} \\ \\ \: \:[/tex]

Now, Subtracting 7 from both sides :

[tex] \\ \longrightarrow \qquad{ {{ \sf{ 7 - 2y - 7=5 - 7 \: \:}}}} \\ \: \:[/tex]

[tex]\longrightarrow \qquad{ {{ \sf{ - 2y = - 2\: \:}}}} \\ \\ \: \:[/tex]

Dividing both sides by -2 we get :

[tex] \\ \longrightarrow \qquad{ {{ \sf{ \frac{ - 2y}{ - 2} = \frac{ - 2}{ - 2} \: \:}}}} \\ \: \:[/tex]

[tex] \longrightarrow \qquad{ \underline{\boxed{ \pmb{ \mathfrak{y = \: \: 1}}}}} \: \bigstar\\ \:

[/tex]

  • Therefore, The value of y is 1

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

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

[tex]\longrightarrow \qquad{ {{ \sf{ x+1=7 \: \:}}}} \\ \\ \: \:[/tex]

Subracting 1 from both sides we get :

[tex] \\ \longrightarrow \qquad{ {{ \sf{ x+1 - 1=7 - 1\: \:}}}} \\ \: \:[/tex]

[tex] \longrightarrow \qquad{ \underline {\boxed{ \pmb{ \mathfrak{x = \: \: 6}}}}} \: \: \bigstar\\ \: [/tex]

  • Therefore, The value of x is 16

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