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QUESTION IMAGE

move each equation into the column that accurately describes the number…

Question

move each equation into the column that accurately describes the number of solutions the equation has.
one solution
no solutions
infinitely many solutions
14x + 10 = 2x + 12x
2(3x + 1) = 3(2x + 4)
3(2x + 3) = 6x + 9
3(x - 1) + 10 = -2x + 12
2x + 5 = 3x - 7

Explanation:

Response

To solve this, we analyze each equation:

Equation 1: \(14x + 10 = 2x + 12x\)

Step 1: Simplify the right - hand side

Combine like terms on the right: \(2x+12x = 14x\). So the equation becomes \(14x + 10=14x\).

Step 2: Subtract \(14x\) from both sides

\(14x - 14x+10=14x - 14x\), which simplifies to \(10 = 0\). This is a contradiction, so the equation has no solutions.

Equation 2: \(2(3x + 1)=3(2x + 4)\)

Step 1: Expand both sides

Using the distributive property \(a(b + c)=ab+ac\), we get \(6x+2 = 6x + 12\).

Step 2: Subtract \(6x\) from both sides

\(6x-6x + 2=6x - 6x+12\), which simplifies to \(2 = 12\). This is a contradiction, so the equation has no solutions.

Equation 3: \(3(2x + 3)=6x + 9\)

Step 1: Expand the left - hand side

Using the distributive property, \(6x+9 = 6x + 9\).

Step 2: Analyze the equation

If we subtract \(6x\) from both sides, we get \(9 = 9\). This is always true, so the equation has infinitely many solutions.

Equation 4: \(2x + 5=3x - 7\)

Step 1: Subtract \(2x\) from both sides

\(2x-2x + 5=3x-2x - 7\), which simplifies to \(5=x - 7\).

Step 2: Add 7 to both sides

\(5 + 7=x-7 + 7\), so \(x = 12\). This equation has one solution.

Equation 5: \(3(x - 1)+10=-2x + 12\)

Step 1: Expand the left - hand side

Using the distributive property: \(3x-3 + 10=-2x + 12\). Combine like terms on the left: \(3x + 7=-2x + 12\).

Step 2: Add \(2x\) to both sides

\(3x+2x + 7=-2x+2x + 12\), which gives \(5x + 7=12\).

Step 3: Subtract 7 from both sides

\(5x+7 - 7=12 - 7\), so \(5x = 5\).

Step 4: Divide by 5

\(x=\frac{5}{5}=1\). This equation has one solution.

Classification:
  • One Solution: \(2x + 5 = 3x-7\), \(3(x - 1)+10=-2x + 12\)
  • No Solutions: \(14x + 10 = 2x+12x\), \(2(3x + 1)=3(2x + 4)\)
  • Infinitely Many Solutions: \(3(2x + 3)=6x + 9\)

If we were to present the final answer in terms of moving the equations:

  • Move \(2x + 5 = 3x-7\) and \(3(x - 1)+10=-2x + 12\) to the "One Solution" column.
  • Move \(14x + 10 = 2x + 12x\) and \(2(3x + 1)=3(2x + 4)\) to the "No Solutions" column.
  • Move \(3(2x + 3)=6x + 9\) to the "Infinitely Many Solutions" column.

Answer:

To solve this, we analyze each equation:

Equation 1: \(14x + 10 = 2x + 12x\)

Step 1: Simplify the right - hand side

Combine like terms on the right: \(2x+12x = 14x\). So the equation becomes \(14x + 10=14x\).

Step 2: Subtract \(14x\) from both sides

\(14x - 14x+10=14x - 14x\), which simplifies to \(10 = 0\). This is a contradiction, so the equation has no solutions.

Equation 2: \(2(3x + 1)=3(2x + 4)\)

Step 1: Expand both sides

Using the distributive property \(a(b + c)=ab+ac\), we get \(6x+2 = 6x + 12\).

Step 2: Subtract \(6x\) from both sides

\(6x-6x + 2=6x - 6x+12\), which simplifies to \(2 = 12\). This is a contradiction, so the equation has no solutions.

Equation 3: \(3(2x + 3)=6x + 9\)

Step 1: Expand the left - hand side

Using the distributive property, \(6x+9 = 6x + 9\).

Step 2: Analyze the equation

If we subtract \(6x\) from both sides, we get \(9 = 9\). This is always true, so the equation has infinitely many solutions.

Equation 4: \(2x + 5=3x - 7\)

Step 1: Subtract \(2x\) from both sides

\(2x-2x + 5=3x-2x - 7\), which simplifies to \(5=x - 7\).

Step 2: Add 7 to both sides

\(5 + 7=x-7 + 7\), so \(x = 12\). This equation has one solution.

Equation 5: \(3(x - 1)+10=-2x + 12\)

Step 1: Expand the left - hand side

Using the distributive property: \(3x-3 + 10=-2x + 12\). Combine like terms on the left: \(3x + 7=-2x + 12\).

Step 2: Add \(2x\) to both sides

\(3x+2x + 7=-2x+2x + 12\), which gives \(5x + 7=12\).

Step 3: Subtract 7 from both sides

\(5x+7 - 7=12 - 7\), so \(5x = 5\).

Step 4: Divide by 5

\(x=\frac{5}{5}=1\). This equation has one solution.

Classification:
  • One Solution: \(2x + 5 = 3x-7\), \(3(x - 1)+10=-2x + 12\)
  • No Solutions: \(14x + 10 = 2x+12x\), \(2(3x + 1)=3(2x + 4)\)
  • Infinitely Many Solutions: \(3(2x + 3)=6x + 9\)

If we were to present the final answer in terms of moving the equations:

  • Move \(2x + 5 = 3x-7\) and \(3(x - 1)+10=-2x + 12\) to the "One Solution" column.
  • Move \(14x + 10 = 2x + 12x\) and \(2(3x + 1)=3(2x + 4)\) to the "No Solutions" column.
  • Move \(3(2x + 3)=6x + 9\) to the "Infinitely Many Solutions" column.