Question

4y - 9 - 6y = 2(y + 5) - 3 solve for y

Answer 1

`\huge{ \boxed{ \bold{ \tt{y = - 4}}}}`

Explanation:

`\sf{4y - 9 - 6y = 2(y + 5) - 3}`

Subtract like terms : 6y from 4y

`\longrightarrow{ \sf{ - 2y - 9 = 2(y + 5) - 3}}`

Distribute 2 through the parentheses

`\longrightarrow{ \sf{ - 2y - 9 = 2 *y + 2 *5 \: - 3}}`

`\longrightarrow{ \sf{ - 2y - 9 = 2y + 10 - 3}}`

Subtract 3 from 10

`\longrightarrow{ \sf{ - 2y - 9 = 2y + 7}}`

Move 2y to left hand side and change it's sign

`\longrightarrow{ \sf{ - 2y - 2y - 9 = 7}}`

Move 9 to right hand side and change it's sign

`\longrightarrow{ \sf{ - 2y - 2y = 7 + 9}}`

Combine like terms

`\longrightarrow{ \sf{ - 4y =7 + 9}}`

Add the numbers : 7 and 9

`\longrightarrow{ \sf{ - 4y = 16}}`

Divide both sides by -4

`\longrightarrow{ \sf{ ( - 4y)/( - 4) = (16)/( - 4)}}`

Remember : Dividing a positive integer by any negative integer gives a negative integer.

`\longrightarrow{ \boxed{ \sf{y = - 4}}}`

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Let's check whether the value of y is -4 or not.

Verification

L.H.S :
`\sf{4y - 9 - 6y}`

plug the value of y and simplify

`\dashrightarrow{ \sf{4 * (- 4 )- 9 - 6 * (- 4)}}`

`\dashrightarrow{ \sf{ - 16 - 9 - ( - 24)}}`

`\dashrightarrow{ \sf{ - 16 - 9 + 24}}`

`\dashrightarrow{ \sf{ - 16 + 15}}`

`\dashrightarrow{ \sf{ - 1}}`

R.H.S :
`\sf{2(y + 5) - 3}`

plug the value of y and simplify

`\dashrightarrow{ \sf{2( - 4 + 5) - 3}}`

`\dashrightarrow{ \sf{2 *1 - 3}}`

`\dashrightarrow{ \sf{2 - 3 }}`

`\dashrightarrow{ \sf{ - 1}}`

∴ L.H.S = R.H.S

The value of y is -4

And we're done!

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Rules for solving an equation

• If any equation contains fractions , multiply each term by the LCM of denominators.
• Remove the brackets , if any.
• Collect the terms with the variable to left hand side and constant terms to the right side by changing their signs ' + ' into ' - ' and ' - ' into ' + '.
• Simplify and get the single term on each side.
• Divide each side by the coefficient of variable and then get the value of variable.

Hope I helped!

Best regards! :D

~TheAnimeGirl