To solve the equation \( x \cdot 4 + y^2 = 48 \), we can rearrange it into the form: \[ 4x + y^2 = 48 \] This means we can isolate one of the variables, say \( y^2 \): \[ y^2 = 48 - 4x \] Now, the right side must be non-negative since \( y^2 \) cannot be negative. Therefore: \[ 48 - 4x \geq 0 \] From this inequality, we can solve for \( x \): \[ 48 \geq 4x \quad \Rightarrow \quad 12 \geq x \quad \Rightarrow \quad x \leq 12 \] Now, substituting integer values for \( x \) from 0 to 12 will give us possible values for \( y^2 \): 1. If \( x = 0 \): \[ y^2 = 48 - 4(0) = 48 \quad \Rightarrow \quad y = \pm \sqrt{48} \quad \Rightarrow \quad y = \pm 4\sqrt{3} \] 2. If \( x = 1 \): \[ y^2 = 48 - 4(1) = 44 \quad \Rightarrow \quad y = \pm \sqrt{44} \quad \Rightarrow \quad y = \pm 2\sqrt{11} \] 3. If \( x = 2 \): \[ y^2 = 48 - 4(2) = 40 \quad \Rightarrow \quad y = \pm \sqrt{40} \quad \Rightarrow \quad y = \pm 2\sqrt{10} \] 4. If \( x = 3 \): \[ y^2 = 48 - 4(3) = 36 \quad \Rightarrow \quad y = \pm \sqrt{36} \quad \Rightarrow \quad y = \pm 6 \] 5. If \( x = 4 \): \[ y^2 = 48 - 4(4) = 32 \quad \Rightarrow \quad y = \pm \sqrt{32} \quad \Rightarrow \quad y = \pm 4\sqrt{2} \] 6. If \( x = 5 \): \[ y^2 = 48 - 4(5) = 28 \quad \Rightarrow \quad y = \pm \sqrt{28} \quad \Rightarrow \quad y = \pm 2\sqrt{7} \] 7. If \( x = 6 \): \[ y^2 = 48 - 4(6) = 24 \quad \Rightarrow \quad y = \pm \sqrt{24} \quad \Rightarrow \quad y = \pm 2\sqrt{6} \] 8. If \( x = 7 \): \[ y^2 = 48 - 4(7) = 20 \quad \Rightarrow \quad y = \pm \sqrt{20} \quad \Rightarrow \quad y = \pm 2\sqrt{5} \] 9. If \( x = 8 \): \[ y^2 = 48 - 4(8) = 16 \quad \Rightarrow \quad y = \pm \sqrt{16} \quad \Rightarrow \quad y = \pm 4 \] 10. If \( x = 9 \): \[ y^2 = 48 - 4(9) = 12 \quad \Rightarrow \quad y = \pm \sqrt{12} \quad \Rightarrow \quad y = \pm 2\sqrt{3} \] 11. If \( x = 10 \): \[ y^2 = 48 - 4(10) = 8 \quad \Rightarrow \quad y = \pm \sqrt{8} \quad \Rightarrow \quad y = \pm 2\sqrt{2} \] 12. If \( x = 11 \): \[ y^2 = 48 - 4(11) = 4 \quad \Rightarrow \quad y = \pm \sqrt{4} \quad \Rightarrow \quad y = \pm 2 \] 13. If \( x = 12 \): \[ y^2 = 48 - 4(12) = 0 \quad \Rightarrow \quad y = 0 \] Thus, we have various pairs of \( (x, y) \) that satisfy the equation. Some notable pairs include: - \( (3, 6) \) - \( (8, 4) \) - \( (11, 2) \) - \( (12, 0) \) Would you like to analyze further or clarify anything?

To solve the equation \( y^2 \cdot x + 3 = 19 \), we can start by isolating the term \( y^2 \cdot x \): \[ y^2 \cdot x = 19 - 3 \] This simplifies to: \[ y^2 \cdot x = 16 \] Next, we can express this as: \[ y^2 = \frac{16}{x} \] Since \( y^2 \) must be non-negative, \( x \) must be a positive value. We can analyze integer values of \( x \) that divide 16 to find corresponding values for \( y \). The possible positive integer values for \( x \) that divide 16 are: \( 1, 2, 4, 8, 16 \). ### Calculating Corresponding \( y \): 1. **If \( x = 1 \)**: \[ y^2 = \frac{16}{1} = 16 \implies y = \pm 4 \] 2. **If \( x = 2 \)**: \[ y^2 = \frac{16}{2} = 8 \implies y = \pm 2\sqrt{2} \] 3. **If \( x = 4 \)**: \[ y^2 = \frac{16}{4} = 4 \implies y = \pm 2 \] 4. **If \( x = 8 \)**: \[ y^2 = \frac{16}{8} = 2 \implies y = \pm \sqrt{2} \] 5. **If \( x = 16 \)**: \[ y^2 = \frac{16}{16} = 1 \implies y = \pm 1 \] ### Summary of Solutions: The pairs \( (x, y) \) that satisfy the equation \( y^2 \cdot x + 3 = 19 \) are: - \( (1, 4) \) and \( (1, -4) \) - \( (2, 2\sqrt{2}) \) and \( (2, -2\sqrt{2}) \) - \( (4, 2) \) and \( (4, -2) \) - \( (8, \sqrt{2}) \) and \( (8, -\sqrt{2}) \) - \( (16, 1) \) and \( (16, -1) \) If you have any further questions or need additional clarification, feel free to ask!

(?) * 4 + ?^2= 48

To solve the equation \( x \cdot 4 + y^2 = 48 \), we can rearrange it into the form:

\[ 4x + y^2 = 48 \]

This means we can isolate one of the variables, say \( y^2 \):

\[ y^2 = 48 - 4x \]

Now, the right side must be non-negative since \( y^2 \) cannot be negative. Therefore:

\[ 48 - 4x \geq 0 \]

From this inequality, we can solve for \( x \):

\[ 48 \geq 4x \quad \Rightarrow \quad 12 \geq x \quad \Rightarrow \quad x \leq 12 \]

Now, substituting integer values for \( x \) from 0 to 12 will give us possible values for \( y^2 \):

If \( x = 0 \): \[ y^2 = 48 - 4(0) = 48 \quad \Rightarrow \quad y = \pm \sqrt{48} \quad \Rightarrow \quad y = \pm 4\sqrt{3} \]
If \( x = 1 \): \[ y^2 = 48 - 4(1) = 44 \quad \Rightarrow \quad y = \pm \sqrt{44} \quad \Rightarrow \quad y = \pm 2\sqrt{11} \]
If \( x = 2 \): \[ y^2 = 48 - 4(2) = 40 \quad \Rightarrow \quad y = \pm \sqrt{40} \quad \Rightarrow \quad y = \pm 2\sqrt{10} \]
If \( x = 3 \): \[ y^2 = 48 - 4(3) = 36 \quad \Rightarrow \quad y = \pm \sqrt{36} \quad \Rightarrow \quad y = \pm 6 \]
If \( x = 4 \): \[ y^2 = 48 - 4(4) = 32 \quad \Rightarrow \quad y = \pm \sqrt{32} \quad \Rightarrow \quad y = \pm 4\sqrt{2} \]
If \( x = 5 \): \[ y^2 = 48 - 4(5) = 28 \quad \Rightarrow \quad y = \pm \sqrt{28} \quad \Rightarrow \quad y = \pm 2\sqrt{7} \]
If \( x = 6 \): \[ y^2 = 48 - 4(6) = 24 \quad \Rightarrow \quad y = \pm \sqrt{24} \quad \Rightarrow \quad y = \pm 2\sqrt{6} \]
If \( x = 7 \): \[ y^2 = 48 - 4(7) = 20 \quad \Rightarrow \quad y = \pm \sqrt{20} \quad \Rightarrow \quad y = \pm 2\sqrt{5} \]
If \( x = 8 \): \[ y^2 = 48 - 4(8) = 16 \quad \Rightarrow \quad y = \pm \sqrt{16} \quad \Rightarrow \quad y = \pm 4 \]
If \( x = 9 \): \[ y^2 = 48 - 4(9) = 12 \quad \Rightarrow \quad y = \pm \sqrt{12} \quad \Rightarrow \quad y = \pm 2\sqrt{3} \]
If \( x = 10 \): \[ y^2 = 48 - 4(10) = 8 \quad \Rightarrow \quad y = \pm \sqrt{8} \quad \Rightarrow \quad y = \pm 2\sqrt{2} \]
If \( x = 11 \): \[ y^2 = 48 - 4(11) = 4 \quad \Rightarrow \quad y = \pm \sqrt{4} \quad \Rightarrow \quad y = \pm 2 \]
If \( x = 12 \): \[ y^2 = 48 - 4(12) = 0 \quad \Rightarrow \quad y = 0 \]

Thus, we have various pairs of \( (x, y) \) that satisfy the equation. Some notable pairs include:

\( (3, 6) \)
\( (8, 4) \)
\( (11, 2) \)
\( (12, 0) \)

Would you like to analyze further or clarify anything?

?^2 * ? + 3= 19

To solve the equation \( y^2 \cdot x + 3 = 19 \), we can start by isolating the term \( y^2 \cdot x \):

\[ y^2 \cdot x = 19 - 3 \]

This simplifies to:

\[ y^2 \cdot x = 16 \]

Next, we can express this as:

\[ y^2 = \frac{16}{x} \]

Since \( y^2 \) must be non-negative, \( x \) must be a positive value. We can analyze integer values of \( x \) that divide 16 to find corresponding values for \( y \).

The possible positive integer values for \( x \) that divide 16 are: \( 1, 2, 4, 8, 16 \).

Calculating Corresponding \( y \):

If \( x = 1 \): \[ y^2 = \frac{16}{1} = 16 \implies y = \pm 4 \]
If \( x = 2 \): \[ y^2 = \frac{16}{2} = 8 \implies y = \pm 2\sqrt{2} \]
If \( x = 4 \): \[ y^2 = \frac{16}{4} = 4 \implies y = \pm 2 \]
If \( x = 8 \): \[ y^2 = \frac{16}{8} = 2 \implies y = \pm \sqrt{2} \]
If \( x = 16 \): \[ y^2 = \frac{16}{16} = 1 \implies y = \pm 1 \]

Summary of Solutions:

The pairs \( (x, y) \) that satisfy the equation \( y^2 \cdot x + 3 = 19 \) are:

\( (1, 4) \) and \( (1, -4) \)
\( (2, 2\sqrt{2}) \) and \( (2, -2\sqrt{2}) \)
\( (4, 2) \) and \( (4, -2) \)
\( (8, \sqrt{2}) \) and \( (8, -\sqrt{2}) \)
\( (16, 1) \) and \( (16, -1) \)

If you have any further questions or need additional clarification, feel free to ask!

? * ? + 4^2 = 28

To solve the equation

\[ x \cdot y + 4^2 = 28, \]

first, we can simplify the equation by calculating \(4^2\):

\[ 4^2 = 16. \]

Substituting this back into the equation gives:

\[ x \cdot y + 16 = 28. \]

Next, we can isolate the product \(x \cdot y\):

\[ x \cdot y = 28 - 16, \]

which simplifies to:

\[ x \cdot y = 12. \]

Now, we need to find pairs of integers \( (x, y) \) such that their product is 12. The integer factor pairs of 12 are:

\( (1, 12) \)
\( (2, 6) \)
\( (3, 4) \)
\( (4, 3) \)
\( (6, 2) \)
\( (12, 1) \)
\( (-1, -12) \)
\( (-2, -6) \)
\( (-3, -4) \)
\( (-4, -3) \)
\( (-6, -2) \)
\( (-12, -1) \)

Thus, the pairs \( (x, y) \) that satisfy the equation \( x \cdot y + 4^2 = 28 \) are:

\( (1, 12) \)
\( (12, 1) \)
\( (2, 6) \)
\( (6, 2) \)
\( (3, 4) \)
\( (4, 3) \)
\( (-1, -12) \)
\( (-12, -1) \)
\( (-2, -6) \)
\( (-6, -2) \)
\( (-3, -4) \)
\( (-4, -3) \)

If you have any further questions or need additional clarification, feel free to ask!

? * ? + 5 = 34

To solve the equation

\[ x \cdot y + 5 = 34, \]

we can start by isolating the product \(x \cdot y\). Subtract 5 from both sides:

\[ x \cdot y = 34 - 5. \]

This simplifies to:

\[ x \cdot y = 29. \]

Now, we want to find integer pairs \( (x, y) \) such that their product is 29. Since 29 is a prime number, its only integer factor pairs are:

\( (1, 29) \)
\( (29, 1) \)
\( (-1, -29) \)
\( (-29, -1) \)

Thus, the pairs \( (x, y) \) that satisfy the equation \( x \cdot y + 5 = 34 \) are:

\( (1, 29) \)
\( (29, 1) \)
\( (-1, -29) \)
\( (-29, -1) \)

If you have any further questions or need clarifications, feel free to ask!

? * 3 + 5 = 34

To solve the equation

\[ x \cdot 3 + 5 = 34, \]

first, we need to isolate the term involving \(x\). We can do this by subtracting 5 from both sides of the equation:

\[ x \cdot 3 = 34 - 5. \]

This simplifies to:

\[ x \cdot 3 = 29. \]

Next, divide both sides by 3 to solve for \(x\):

\[ x = \frac{29}{3}. \]

Therefore, the solution is:

\[ x = \frac{29}{3} \approx 9.67. \]

If you were looking for an integer solution, there is none since 29 is not divisible by 3 evenly. However, if you require further assistance or clarifications, please let me know!

try again its a number 1-7

Let's solve the equation

\[ ? \cdot 3 + 5 = 34 \]

We want to find a value for ? (or \(x\)) in the range of 1 to 7.

Starting from the equation:

Subtract 5 from both sides: \[ ? \cdot 3 = 34 - 5 \] \[ ? \cdot 3 = 29 \]
Now divide both sides by 3: \[ ? = \frac{29}{3} \] \[ ? \approx 9.67 \]

Since \(9.67\) is not in the range \(1-7\), let's take a step back and check if we're looking for a number \(x\) between 1 and 7 instead, or if we should solve and find all potential values before confirming they fit in the specified range.

Since you specified a number between 1 and 7, let's check what integer values from 1 to 7 would yield a true statement in the equation.

We'll check the integers \(1, 2, 3, 4, 5, 6, 7\):

For \(x = 1\): \[ 1 \cdot 3 + 5 = 3 + 5 = 8 \]
For \(x = 2\): \[ 2 \cdot 3 + 5 = 6 + 5 = 11 \]
For \(x = 3\): \[ 3 \cdot 3 + 5 = 9 + 5 = 14 \]
For \(x = 4\): \[ 4 \cdot 3 + 5 = 12 + 5 = 17 \]
For \(x = 5\): \[ 5 \cdot 3 + 5 = 15 + 5 = 20 \]
For \(x = 6\): \[ 6 \cdot 3 + 5 = 18 + 5 = 23 \]
For \(x = 7\): \[ 7 \cdot 3 + 5 = 21 + 5 = 26 \]

None of the values from 1 to 7 satisfy the equation \(x \cdot 3 + 5 = 34\).

It seems there may be a misunderstanding with the parameters given. Please clarify if you meant a different type of condition or approach!