using inverse operations to solve 5x -7 =3 show all steps for getting the variable by itself then check your answer by plugging it back into the original equations
To solve the equation 5x - 7 = 3, we need to isolate the variable x.
Step 1: Start by getting rid of the constant term -7 on the left side of the equation. We can do this by adding 7 to both sides of the equation.
5x - 7 + 7 = 3 + 7
5x = 10
Step 2: Now the equation is simplified to 5x = 10. To further isolate x, divide both sides of the equation by the coefficient 5.
5x/5 = 10/5
x = 2
Hence, we found that x = 2 after simplifying the equation using inverse operations.
To verify our answer, we can substitute x = 2 back into the original equation 5x - 7 = 3 and see if it holds true.
5(2) - 7 = 3
10 - 7 = 3
3 = 3
Since both sides of the equation are equal, we can conclude that our solution x = 2 is correct.
To solve the equation 5x - 7 = 3 using inverse operations to isolate the variable x, follow these steps:
Step 1: Add 7 to both sides of the equation to get rid of the -7:
5x - 7 + 7 = 3 + 7
5x = 10
Step 2: Divide both sides of the equation by 5 to get x alone:
(5x) / 5 = 10 / 5
x = 2
Therefore, the solution to the equation is x = 2.
To check our answer, we will substitute x = 2 back into the original equation:
5x - 7 = 3
Using x = 2:
5(2) - 7 = 3
10 - 7 = 3
3 = 3
Since both sides of the equation are equal, our solution x = 2 is correct.
To solve the equation 5x - 7 = 3 using inverse operations, we want to isolate the variable x by performing the opposite operations step-by-step.
Step 1: Add 7 to both sides of the equation to get rid of the constant term on the left side.
5x - 7 + 7 = 3 + 7
5x = 10
Step 2: Divide both sides of the equation by 5 to undo the multiplication of 5 on the variable x.
(5x)/5 = 10/5
x = 2
Now, let's check our solution by plugging the value of x back into the original equation.
Original equation:
5x - 7 = 3
Plugging in x = 2:
5(2) - 7 = 3
10 - 7 = 3
3 = 3
Since the equation is true, we can conclude that x = 2 is the correct solution.