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galben [10]
3 years ago
9

Which of the following equations have infinitely many solutions?

Mathematics
1 answer:
olga55 [171]3 years ago
8 0

The  following equations have infinitely many solutions:

6(2x + 4) = 3(4x + 8) ⇒ I

0.5(8x + 4) = 4x + 2 ⇒ II

2(6x + 4) = 4(3x + 2) ⇒ V

The answer is l, ll, and V ⇒ A

Step-by-step explanation:

The equation has infinitely many solutions if:

  • The variable is disappeared
  • Then the numerical terms are equal

I.

∵ 6(2x + 4) = 3(4x + 8)

- Simplify the two sides

∵ 6(2x + 4) = 6(2x) + 6(4) = 12x + 24

∵ 3(4x + 8) = 3(4x) + 3(8) = 12x + 24

∴ 12x + 24 = 12x + 24

- Subtract 12x from both sides

∴ 24 = 24

- The variable disappear and two sides are equal

∴ The equation has infinitely many solutions

II.

∵ 0.5(8x + 4) = 4x + 2

- Simplify the left hand side

∵ 0.5(8x + 4) = 0.5(8x) + 0.5(4) = 4x + 2

∵ The right hand side is 4x + 2

∴ 4x + 2 = 4x + 2

- Subtract 4x from both sides

∴ 2 = 2

- The variable disappear and two sides are equal

∴ The equation has infinitely many solutions

III.

∵ 7x + 8 = 7(x + 8)

- Simplify the right hand side

∵ 7(x + 8) = 7(x) + 7(8) = 7x + 56

∴ 7x + 8 = 7x + 56

- Subtract 7x from both sides

∴ 8 = 56 ⇒ it can not be

∴ L.H.S ≠ R.H.S

∴ The equation has no solution

IV.

∵ 3x + 4 = 7x - 2

- Subtract 3x from both sides

∴ 4 = 4x - 2

- Add 2 to both sides

∴ 6 = 4x

- Divide both sides by 4

∴ 1.5 = x

∴ The equation has one solution

V.

∵ 2(6x + 4) = 4(3x + 2)

- Simplify the both sides

∵ 2(6x + 4) = 2(6x) + 2(4) = 12x + 8

∵ 4(3x + 2) = 4(3x) + 4(2) = 12x + 8

∴ 12x + 8 = 12x + 8

- Subtract 12x from both sides

∴ 8 = 8

- The variable disappear and two sides are equal

∴ The equation has infinitely many solutions

The  following equations have infinitely many solutions:

6(2x + 4) = 3(4x + 8) ⇒ I

0.5(8x + 4) = 4x + 2 ⇒ II

2(6x + 4) = 4(3x + 2) ⇒ V

Learn more:

You can learn more about the solutions of the linear equations in

brainly.com/question/6075514

#LearnwithBrainly

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Given that (ax^2 + bx + 3) (x + d) = x^3 + 6x^2 + 11x + 12<br> a + 2b - d = ?
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Answer:

Let's solve for a.

(ax2+bx+3)(x+d)=x3+6x2+11x+12a+2b−d

Step 1: Add -12a to both sides.

adx2+ax3+bdx+bx2+3d+3x+−12a=x3+6x2+12a+2b−d+11x+−12a

adx2+ax3+bdx+bx2−12a+3d+3x=x3+6x2+2b−d+11x

Step 2: Add -bdx to both sides.

adx2+ax3+bdx+bx2−12a+3d+3x+−bdx=x3+6x2+2b−d+11x+−bdx

adx2+ax3+bx2−12a+3d+3x=−bdx+x3+6x2+2b−d+11x

Step 3: Add -bx^2 to both sides.

adx2+ax3+bx2−12a+3d+3x+−bx2=−bdx+x3+6x2+2b−d+11x+−bx2

adx2+ax3−12a+3d+3x=−bdx−bx2+x3+6x2+2b−d+11x

Step 4: Add -3d to both sides.

adx2+ax3−12a+3d+3x+−3d=−bdx−bx2+x3+6x2+2b−d+11x+−3d

adx2+ax3−12a+3x=−bdx−bx2+x3+6x2+2b−4d+11x

Step 5: Add -3x to both sides.

adx2+ax3−12a+3x+−3x=−bdx−bx2+x3+6x2+2b−4d+11x+−3x

adx2+ax3−12a=−bdx−bx2+x3+6x2+2b−4d+8x

Step 6: Factor out variable a.

a(dx2+x3−12)=−bdx−bx2+x3+6x2+2b−4d+8x

Step 7: Divide both sides by dx^2+x^3-12.

a(dx2+x3−12)

dx2+x3−12

=

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

a=

−bdx−bx2+x3+6x2+2b−4d+8x

dx2+x3−12

Answer:

a=

−bdx−bx2+x3+6x2+2b−4d+8x/

dx2+x3−12

Step-by-step explanation:

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