Hey ! there
Answer:
- Value of x in equation A is 2 .
- Value of x in equation B is 8 .
Step-by-step explanation:
In this question we are provided with two equations that are 17x + 12 = 54 - 4x and 5x - 10 + 2x = 6(x + 1) - x . And we are asked to solve the equation that means we have to find the value of x for both the equations .
Solution : -
A)
[tex] \longmapsto \qquad \: 17x + 12 = 54 - 4x[/tex]
Step 1 : Subtracting 12 on both sides :
[tex] \longmapsto \qquad \: 17x + \cancel{ 12} - \cancel{12} = \bold{54} - 4x - \bold{ 12}[/tex]
Simplifying it ,
[tex] \longmapsto \qquad \: 17x = 42 - 4x[/tex]
Step 2 : Adding 4x on both sides :
[tex] \longmapsto \qquad \: 17x + 4x = 42 - \cancel{4x} + \cancel{4x}[/tex]
On further calculations, we get :
[tex] \longmapsto \qquad \: 21x = 42[/tex]
Step 3 : Dividing with 21 on both sides :
[tex] \longmapsto \qquad \: \dfrac{ \cancel{21}x}{ \cancel{21} } = \cancel{ \dfrac{42}{21} }[/tex]
We get :
[tex] \longmapsto \qquad \: \red{\underline{\boxed{ \frak{x = 2}}}} \quad \bigstar[/tex]
- Henceforth , value of x is 2 .
Verifying : -
Now , we are verifying our answer by substituting value of x as 2 in the given equation . So ,
- 17 ( 2 ) + 12 = 54 - 4 ( 2 )
Therefore , our solution is correct .
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B)
[tex] \longmapsto \qquad \: 5x - 10 + 2x = 6(x + 1) - x[/tex]
Step 1 : Adding like terms on left side :
[tex] \longmapsto \qquad \: 7x - 10 = 6(x + 1) - x[/tex]
Step 2 : Solving parenthesis on right side by using distributive property . ( Multiplying 6 with x and 1 both ) :
[tex] \longmapsto \qquad \: 7x - 10 = \bold{6x} + 6 - \bold{x}[/tex]
Step 3 : Solving like terms on right side :
[tex] \longmapsto \qquad \: 7x - 10 = 5x + 6[/tex]
Step 4 : Adding 10 on both sides :
[tex] \longmapsto \qquad \: 7x - \cancel{10} + \cancel{ 10} = 5x + 6 + 10[/tex]
Simplifying it ,
[tex] \longmapsto \qquad \: 7x = 5x + 16[/tex]
Step 5 : Subtracting 5x on both sides :
[tex] \longmapsto \qquad \: 7x - 5x = \cancel{5x }+ 16 - \cancel{5x}[/tex]
On further calculations , we get :
[tex] \longmapsto \qquad \: 2x = 16[/tex]
Step 6 : Dividing with 2 on both sides :
[tex] \longmapsto \qquad \: \dfrac{ \cancel{2}x}{ \cancel{2}} = \cancel{\dfrac{16}{2} }[/tex]
We get :
[tex] \longmapsto \qquad \: \red{\underline{\boxed{ \frak{x = 8}}} } \quad \bigstar[/tex]
- Henceforth , value of x is 8 .
Verifying : -
Now , we are verifying our answer by substituting value of x as 8 in the given equation . So ,
- 5x - 10 + 2x = 6(x + 1) - x
- 5 ( 8 ) - 10 + 2( 8 ) = 6 ( 8 + 1 ) - 8
- 40 ( - 10 ) + 16 = 6 ( 9 ) - 8
Therefore , our solution is correct.
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