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Sagot :
Answer:
- 5 is the value which makes the equation true .
Step-by-step explanation:
In this question we have provided an equation that is 9 ( 3x - 16 ) + 15 = 6x - 24 . And we are asked to write the steps to solve the equation with explanation and to find the value of X .
Solution : -
[tex] \longmapsto \quad \: 9(3x - 16) + 15 = 6x - 24[/tex]
Step 1 : Solving parenthesis on left side using distributive property which means multiplying 9 with 3x as well as -16 :
[tex] \longmapsto \quad \:27x - \bold{144 }+ \bold{15 }= 6x - 24[/tex]
Step 2 : Solving like terms on left side that are -144 and 15 :
[tex] \longmapsto \quad \:27x -129 = 6x - 24[/tex]
Step 3 : Adding 129 on both sides :
[tex] \longmapsto \quad \:27x - \cancel{129} - \cancel{129} = 6x \bold{ - 24 } + \bold{129}[/tex]
Now on cancelling -129 with 129 on left side and solving the terms that are -24 and 129 on right side , We get :
[tex] \longmapsto \quad \:27x = 6x + 105[/tex]
Step 4 : Subtracting with 6x on both sides :
[tex] \longmapsto \quad \: \bold{27x} - \bold{6x} = \cancel{6x} +105 - \cancel{ 6x}[/tex]
On calculating further, We get :
[tex] \longmapsto \quad \:21x = 105[/tex]
Step 5 : Now we are Dividing with 21 on both sides so that we can isolate the variable that is x :
[tex] \longmapsto \quad \: \dfrac{ \cancel{21}x}{ \cancel{21}} = \dfrac{105}{ 21} [/tex]
Now , by cancelling 21 with 21 on left side , We get :
[tex] \longmapsto \quad \:x = \cancel{\dfrac{105}{21}} [/tex]
Step 6 : Now our final step is to simplify the value of x that is 105/21 . We know that 21 × 5 is equal to 105 . So :
[tex] \longmapsto \quad \: \purple{\underline{\boxed{\frak{ x = 5 }}}}[/tex]
- Henceforth , value of x is 5
Verifying : -
Now we are verifying our answer by substituting value of x in the given equation . So ,
- 9 ( 3x - 16 ) + 15 = 6x - 24
- 9 [ 3 ( 5 ) - 16 ] + 15 = 6 ( 5 ) - 24
- 9 ( 15 - 16 ) + 15 = 30 - 24
- 9 ( -1 ) + 15 = 6
- -9 + 15 = 6
- 6 = 6
- L.H.S = R.H.S
- Hence , Verified .
Therefore, our value for x is correct that means it'll makes the equation true .
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