Asked by Makayla
Given the equation 5+x−12=x−7.
Part A: Solve the equation 5+x−12=x−7. In your final answer, be sure to state the solution and include all of your work.
Part B: Use the values x=−4,0,5 to verify your solution to the equation 5+x−12=x−7. In your final answer, include all of your calculations.
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Part A — Solve 5 + x − 12 = x − 7
1. Combine like terms on the left: 5 − 12 = −7, so the equation becomes
x − 7 = x − 7.
2. Subtract x from both sides:
(x − 7) − x = (x − 7) − x → −7 = −7.
Since the resulting statement −7 = −7 is always true and does not restrict x, the original equation is an identity.
Solution: Every real number x satisfies the equation (infinitely many solutions).
Part B — Verify with x = −4, 0, 5
1. x = −4:
Left: 5 + (−4) − 12 = 5 − 4 − 12 = −11.
Right: (−4) − 7 = −11.
Left = Right → checks.
2. x = 0:
Left: 5 + 0 − 12 = 5 − 12 = −7.
Right: 0 − 7 = −7.
Left = Right → checks.
3. x = 5:
Left: 5 + 5 − 12 = 10 − 12 = −2.
Right: 5 − 7 = −2.
Left = Right → checks.
All three test values satisfy the equation, consistent with the conclusion that every real x is a solution.
1. Combine like terms on the left: 5 − 12 = −7, so the equation becomes
x − 7 = x − 7.
2. Subtract x from both sides:
(x − 7) − x = (x − 7) − x → −7 = −7.
Since the resulting statement −7 = −7 is always true and does not restrict x, the original equation is an identity.
Solution: Every real number x satisfies the equation (infinitely many solutions).
Part B — Verify with x = −4, 0, 5
1. x = −4:
Left: 5 + (−4) − 12 = 5 − 4 − 12 = −11.
Right: (−4) − 7 = −11.
Left = Right → checks.
2. x = 0:
Left: 5 + 0 − 12 = 5 − 12 = −7.
Right: 0 − 7 = −7.
Left = Right → checks.
3. x = 5:
Left: 5 + 5 − 12 = 10 − 12 = −2.
Right: 5 − 7 = −2.
Left = Right → checks.
All three test values satisfy the equation, consistent with the conclusion that every real x is a solution.
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