Write Equation: First, let's write down the equation we need to solve: 600x = 7x! - 32. We need to isolate x on one side of the equation to solve for it.Isolate x: Since the equation involves a factorial, we should look for a value of x that makes x! divisible by 7, as this will simplify the equation. We can start by testing small integer values of x.Test Small Values: Let's test x = 1: 600(1) = 7(1)! - 32, which simplifies to 600 = 7 - 32. This is not true, so x = 1 is not the solution.Rearrange Equation: Let's test x = 2: 600(2) = 7(2)! - 32, which simplifies to 1200 = 7(2) - 32. This simplifies to 1200 = 14 - 32, which is not true, so x = 2 is not the solution.Divide by x: Let's test x = 3: 600(3) = 7(3)! - 32, which simplifies to 1800 = 7(6) - 32. This simplifies to 1800 = 42 - 32, which is not true, so x = 3 is not the solution.Find Integer Solution: Let's test x = 4: 600(4) = 7(4)! - 32, which simplifies to 2400 = 7(24) - 32. This simplifies to 2400 = 168 - 32, which is not true, so x = 4 is not the solution.Test Factors of 32: Let's test x = 5: 600(5) = 7(5)! - 32, which simplifies to 3000 = 7(120) - 32. This simplifies to 3000 = 840 - 32, which is not true, so x = 5 is not the solution.No Integer Solution: Let's test x = 6: 600(6) = 7(6)! - 32, which simplifies to 3600 = 7(720) - 32. This simplifies to 3600 = 5040 - 32, which is not true, so x = 6 is not the solution.Let's test x = 7: 600(7) = 7(7)! - 32, which simplifies to 4200 = 7(5040) - 32. This simplifies to 4200 = 35280 - 32, which is not true, so x = 7 is not the solution.Let's test x = 8: 600(8) = 7(8)! - 32, which simplifies to 4800 = 7(40320) - 32. This simplifies to 4800 = 282240 - 32, which is not true, so x = 8 is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving x to one side: 7x! = 600x + 32. Now we can divide both sides by x to simplify the equation, assuming x is not zero.After dividing by x, we get 7(x-1)! = 600 + 32/x. Now we look for an integer solution for x that satisfies this equation.Since (x-1)! must be an integer, 32/x must also be an integer for the right side of the equation to be an integer. This means x must be a factor of 32.The factors of 32 are 1, 2, 4, 8, 16, and 32. We have already tested values of x from 1 to 8 and found that they do not work. Let's test x = 16: 7(16-1)! = 600 + 32/16.This simplifies to 7(15)! = 600 + 2. However, (15)! is a very large number, and adding 2 to 600 will not make it equal to 7 times a factorial. Therefore, x = 16 is not the solution.Finally, let's test x = 32: 7(32-1)! = 600 + 32/32, which simplifies to 7(31)! = 600 + 1. Again, (31)! is a very large number, and adding 1 to 600 will not make it equal to 7 times a factorial. Therefore, x = 32 is not the solution.It appears that there is no integer solution for x in the equation 600x = 7x! - 32. The factorial grows too quickly for the linear term 600x to catch up, and the constant -32 does not significantly affect the outcome.

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$600x = 7x! - 32$

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Algebra 1

Solve linear equations with variables on both sides

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600x = 7x! - 32

Write Equation: First, let's write down the equation we need to solve: $600x = 7x! - 32$ . We need to isolate $x$ on one side of the equation to solve for it.
Isolate $x$ : Since the equation involves a factorial, we should look for a value of $x$ that makes $x!$ divisible by $7$ , as this will simplify the equation. We can start by testing small integer values of $x$ .
Test Small Values: Let's test $x = 1$ : $600(1) = 7(1)! - 32$ , which simplifies to $600 = 7 - 32$ . This is not true, so $x = 1$ is not the solution.
Rearrange Equation: Let's test $x = 2$ : $600(2) = 7(2)! - 32$ , which simplifies to $1200 = 7(2) - 32$ . This simplifies to $1200 = 14 - 32$ , which is not true, so $x = 2$ is not the solution.
Divide by x: Let's test $x = 3$ : $600(3) = 7(3)! - 32$ , which simplifies to $1800 = 7(6) - 32$ . This simplifies to $1800 = 42 - 32$ , which is not true, so $x = 3$ is not the solution.
Find Integer Solution: Let's test $x = 4$ : $600(4) = 7(4)! - 32$ , which simplifies to $2400 = 7(24) - 32$ . This simplifies to $2400 = 168 - 32$ , which is not true, so $x = 4$ is not the solution.
Test Factors of $32$ : Let's test $x = 5$ : $600(5) = 7(5)! - 32$ , which simplifies to $3000 = 7(120) - 32$ . This simplifies to $3000 = 840 - 32$ , which is not true, so $x = 5$ is not the solution.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving $600(6) = 7(6)! - 32$ $5$ to one side: $600(6) = 7(6)! - 32$ $6$ . Now we can divide both sides by $600(6) = 7(6)! - 32$ $5$ to simplify the equation, assuming $600(6) = 7(6)! - 32$ $5$ is not zero.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving $600(6) = 7(6)! - 32$ $5$ to one side: $600(6) = 7(6)! - 32$ $6$ . Now we can divide both sides by $600(6) = 7(6)! - 32$ $5$ to simplify the equation, assuming $600(6) = 7(6)! - 32$ $5$ is not zero.After dividing by $600(6) = 7(6)! - 32$ $5$ , we get $3600 = 7(720) - 32$ $0$ . Now we look for an integer solution for $600(6) = 7(6)! - 32$ $5$ that satisfies this equation.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving $600(6) = 7(6)! - 32$ $5$ to one side: $600(6) = 7(6)! - 32$ $6$ . Now we can divide both sides by $600(6) = 7(6)! - 32$ $5$ to simplify the equation, assuming $600(6) = 7(6)! - 32$ $5$ is not zero.After dividing by $600(6) = 7(6)! - 32$ $5$ , we get $3600 = 7(720) - 32$ $0$ . Now we look for an integer solution for $600(6) = 7(6)! - 32$ $5$ that satisfies this equation.Since $3600 = 7(720) - 32$ $2$ must be an integer, $3600 = 7(720) - 32$ $3$ must also be an integer for the right side of the equation to be an integer. This means $600(6) = 7(6)! - 32$ $5$ must be a factor of $3600 = 7(720) - 32$ $5$ .
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving $600(6) = 7(6)! - 32$ $5$ to one side: $600(6) = 7(6)! - 32$ $6$ . Now we can divide both sides by $600(6) = 7(6)! - 32$ $5$ to simplify the equation, assuming $600(6) = 7(6)! - 32$ $5$ is not zero.After dividing by $600(6) = 7(6)! - 32$ $5$ , we get $3600 = 7(720) - 32$ $0$ . Now we look for an integer solution for $600(6) = 7(6)! - 32$ $5$ that satisfies this equation.Since $3600 = 7(720) - 32$ $2$ must be an integer, $3600 = 7(720) - 32$ $3$ must also be an integer for the right side of the equation to be an integer. This means $600(6) = 7(6)! - 32$ $5$ must be a factor of $3600 = 7(720) - 32$ $5$ .The factors of $3600 = 7(720) - 32$ $5$ are $3600 = 7(720) - 32$ $7$ and $3600 = 7(720) - 32$ $5$ . We have already tested values of $600(6) = 7(6)! - 32$ $5$ from $3600 = 5040 - 32$ $0$ to $3600 = 5040 - 32$ $1$ and found that they do not work. Let's test $3600 = 5040 - 32$ $2$ : $3600 = 5040 - 32$ $3$ .
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving $600(6) = 7(6)! - 32$ $5$ to one side: $600(6) = 7(6)! - 32$ $6$ . Now we can divide both sides by $600(6) = 7(6)! - 32$ $5$ to simplify the equation, assuming $600(6) = 7(6)! - 32$ $5$ is not zero.After dividing by $600(6) = 7(6)! - 32$ $5$ , we get $3600 = 7(720) - 32$ $0$ . Now we look for an integer solution for $600(6) = 7(6)! - 32$ $5$ that satisfies this equation.Since $3600 = 7(720) - 32$ $2$ must be an integer, $3600 = 7(720) - 32$ $3$ must also be an integer for the right side of the equation to be an integer. This means $600(6) = 7(6)! - 32$ $5$ must be a factor of $3600 = 7(720) - 32$ $5$ .The factors of $3600 = 7(720) - 32$ $5$ are $3600 = 7(720) - 32$ $7$ and $3600 = 7(720) - 32$ $5$ . We have already tested values of $600(6) = 7(6)! - 32$ $5$ from $3600 = 5040 - 32$ $0$ to $3600 = 5040 - 32$ $1$ and found that they do not work. Let's test $3600 = 5040 - 32$ $2$ : $3600 = 5040 - 32$ $3$ .This simplifies to $3600 = 5040 - 32$ $4$ . However, $3600 = 5040 - 32$ $5$ is a very large number, and adding $3600 = 5040 - 32$ $6$ to $3600 = 5040 - 32$ $7$ will not make it equal to $3600 = 5040 - 32$ $8$ times a factorial. Therefore, $3600 = 5040 - 32$ $2$ is not the solution.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving $600(6) = 7(6)! - 32$ $5$ to one side: $600(6) = 7(6)! - 32$ $6$ . Now we can divide both sides by $600(6) = 7(6)! - 32$ $5$ to simplify the equation, assuming $600(6) = 7(6)! - 32$ $5$ is not zero.After dividing by $600(6) = 7(6)! - 32$ $5$ , we get $3600 = 7(720) - 32$ $0$ . Now we look for an integer solution for $600(6) = 7(6)! - 32$ $5$ that satisfies this equation.Since $3600 = 7(720) - 32$ $2$ must be an integer, $3600 = 7(720) - 32$ $3$ must also be an integer for the right side of the equation to be an integer. This means $600(6) = 7(6)! - 32$ $5$ must be a factor of $3600 = 7(720) - 32$ $5$ .The factors of $3600 = 7(720) - 32$ $5$ are $3600 = 7(720) - 32$ $7$ and $3600 = 7(720) - 32$ $5$ . We have already tested values of $600(6) = 7(6)! - 32$ $5$ from $3600 = 5040 - 32$ $0$ to $3600 = 5040 - 32$ $1$ and found that they do not work. Let's test $3600 = 5040 - 32$ $2$ : $3600 = 5040 - 32$ $3$ .This simplifies to $3600 = 5040 - 32$ $4$ . However, $3600 = 5040 - 32$ $5$ is a very large number, and adding $3600 = 5040 - 32$ $6$ to $3600 = 5040 - 32$ $7$ will not make it equal to $3600 = 5040 - 32$ $8$ times a factorial. Therefore, $3600 = 5040 - 32$ $2$ is not the solution.Finally, let's test $x = 6$ $0$ : $x = 6$ $1$ , which simplifies to $x = 6$ $2$ . Again, $x = 6$ $3$ is a very large number, and adding $3600 = 5040 - 32$ $0$ to $3600 = 5040 - 32$ $7$ will not make it equal to $3600 = 5040 - 32$ $8$ times a factorial. Therefore, $x = 6$ $0$ is not the solution.
No Integer Solution: Let's test $x = 6$ : $600(6) = 7(6)! - 32$ , which simplifies to $3600 = 7(720) - 32$ . This simplifies to $3600 = 5040 - 32$ , which is not true, so $x = 6$ is not the solution.Let's test $x = 7$ : $600(7) = 7(7)! - 32$ , which simplifies to $4200 = 7(5040) - 32$ . This simplifies to $4200 = 35280 - 32$ , which is not true, so $x = 7$ is not the solution.Let's test $600(6) = 7(6)! - 32$ $0$ : $600(6) = 7(6)! - 32$ $1$ , which simplifies to $600(6) = 7(6)! - 32$ $2$ . This simplifies to $600(6) = 7(6)! - 32$ $3$ , which is not true, so $600(6) = 7(6)! - 32$ $0$ is not the solution.It seems that testing individual values is not leading us to a solution, and the factorial grows very quickly, making it impractical to continue this way. We need to find a different approach to solve the equation.Let's rearrange the equation to bring all terms involving $600(6) = 7(6)! - 32$ $5$ to one side: $600(6) = 7(6)! - 32$ $6$ . Now we can divide both sides by $600(6) = 7(6)! - 32$ $5$ to simplify the equation, assuming $600(6) = 7(6)! - 32$ $5$ is not zero.After dividing by $600(6) = 7(6)! - 32$ $5$ , we get $3600 = 7(720) - 32$ $0$ . Now we look for an integer solution for $600(6) = 7(6)! - 32$ $5$ that satisfies this equation.Since $3600 = 7(720) - 32$ $2$ must be an integer, $3600 = 7(720) - 32$ $3$ must also be an integer for the right side of the equation to be an integer. This means $600(6) = 7(6)! - 32$ $5$ must be a factor of $3600 = 7(720) - 32$ $5$ .The factors of $3600 = 7(720) - 32$ $5$ are $3600 = 7(720) - 32$ $7$ and $3600 = 7(720) - 32$ $5$ . We have already tested values of $600(6) = 7(6)! - 32$ $5$ from $3600 = 5040 - 32$ $0$ to $3600 = 5040 - 32$ $1$ and found that they do not work. Let's test $3600 = 5040 - 32$ $2$ : $3600 = 5040 - 32$ $3$ .This simplifies to $3600 = 5040 - 32$ $4$ . However, $3600 = 5040 - 32$ $5$ is a very large number, and adding $3600 = 5040 - 32$ $6$ to $3600 = 5040 - 32$ $7$ will not make it equal to $3600 = 5040 - 32$ $8$ times a factorial. Therefore, $3600 = 5040 - 32$ $2$ is not the solution.Finally, let's test $x = 6$ $0$ : $x = 6$ $1$ , which simplifies to $x = 6$ $2$ . Again, $x = 6$ $3$ is a very large number, and adding $3600 = 5040 - 32$ $0$ to $3600 = 5040 - 32$ $7$ will not make it equal to $3600 = 5040 - 32$ $8$ times a factorial. Therefore, $x = 6$ $0$ is not the solution.It appears that there is no integer solution for $600(6) = 7(6)! - 32$ $5$ in the equation $x = 6$ $9$ . The factorial grows too quickly for the linear term $x = 7$ $0$ to catch up, and the constant $x = 7$ $1$ does not significantly affect the outcome.

$600x = 7x! - 32$

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600x=7x!−32600x = 7x! - 32600x=7x!−32

$600x = 7x! - 32$