Question

4y[3y-7]- 4 (y-3)+24
solve this equation
as soon as possible.​

Answer 1

Answer:

4y[3y-7]- 4 (y-3)+24

=4y×3y-7×4y-4y-12+24

=12y²-28y-4y+12

=12y²-24y+12

12y²-24y+12=0

y²-2y+1  = 0

y²-1y-1y+1 = 0

y(y-1)-1(y-1)=0

(y-1)(y-1)  = 0

factorisation (y-1)²

y=1

Answer 2

Answer :-

• The required number is 12y² - 32y + 36

Step-by-step explanation ::

To Find :-

• Solve the equation :-

4y ( 3y - 7 ) - 4 ( y - 3 ) + 24

Solution :-

$\bf{ \hookrightarrow \: 4y(3y-7)-4(y-3)+24} \\ \\ \\ \sf{ \hookrightarrow \: (4y)(3y) + (4y)(-7) + (-4)(y) + ( - 4)(-3)+24} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + (4y)( - 7) + (-4)(y) + ( - 4)(-3)+24} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y) + (-4)(y) + ( - 4)(-3)+24} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y) + ( - 4y) + ( - 4)(-3)+24} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y) + ( - 4y) + 12 +24} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y + - 4y) + (12 +24)} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} - 32y + (12 +24)} \\ \\ \\ \bf{ \hookrightarrow \: {12y}^{2} - 32y + 36}$

• The required number is 12y² - 32y + 36.

Now, Verification

$\sf{ \hookrightarrow \: 4y(3y-7)-4(y-3)+24 = {12y}^{2} - 32y + 36} \\ \\ \\ \sf{ \hookrightarrow \: (4y)(3y) + (4y)(-7) + (-4)(y) + ( - 4)(-3)+24 = {12y}^{2} - 32y + 36} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + (4y)( - 7) + (-4)(y) + ( - 4)(-3)+24 = {12y}^{2} - 32y + 36} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y) + (-4)(y) + ( - 4)(-3)+24 = {12y}^{2} - 32y + 36} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y) + ( - 4y) + ( - 4)(-3)+24 = {12y}^{2} - 32y + 36} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y) + ( - 4y) + 12 +24 = {12y}^{2} - 32y + 36} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} + ( - 28y + - 4y) + (12 +24) = {12y}^{2} - 32y + 36} \\ \\ \\ \sf{ \hookrightarrow \: {12y}^{2} - 32y + (12 +24) = {12y}^{2} - 32y + 36} \\ \\ \\ \bf{ \hookrightarrow \: {12y}^{2} - 32y + 36 = {12y}^{2} - 32y + 36}$

L.H.S = R.H.S

Hence, Verified!