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examples 6 and 7 solve each equation and state whether the equation has…

Question

examples 6 and 7
solve each equation and state whether the equation has one solution, no solution, or is an identity.

  1. -6y - 3 = 3 - 6y
  2. $\frac{1}{2}(x + 6)=\frac{1}{2}x - 9$
  3. 8q + 12 = 4(3 + 2q)
  4. 21(x + 1) - 6x = 15x + 21
  5. 12y + 48 - 4y = 8(y - 6)
  6. 8(z + 6) = 4(2z + 12)
  7. 2a + 2 = 3(a + 2)
  8. $\frac{1}{4}x + 5=\frac{1}{4}x$
  9. 7(c + 9) = 7c + 63
  10. 4k + 3 = $\frac{1}{4}(8k + 16)$
  11. 3b - 13 + 4b = 7b + 1
  12. $\frac{1}{2}(\frac{1}{2}m - 8)=\frac{1}{4}(m - 16)$

Explanation:

Step1: Solve equation 25

-6y - 3 = 3 - 6y
Add 6y to both sides: -6y+6y - 3 = 3 - 6y+6y
-3 = 3, which is false. So no solution.

Step2: Solve equation 26

\frac{1}{2}(x + 6)=\frac{1}{2}x - 9
Distribute the \frac{1}{2}: \frac{1}{2}x+3=\frac{1}{2}x - 9
Subtract \frac{1}{2}x from both sides: \frac{1}{2}x-\frac{1}{2}x + 3=\frac{1}{2}x-\frac{1}{2}x - 9
3=-9, which is false. So no solution.

Step3: Solve equation 27

8q + 12 = 4(3 + 2q)
Distribute the 4: 8q + 12 = 12+8q
Subtract 8q from both sides: 8q-8q + 12 = 12+8q-8q
12 = 12, which is true for all q. So identity.

Step4: Solve equation 28

21(x + 1)-6x = 15x + 21
Distribute the 21: 21x+21 - 6x = 15x + 21
Combine like - terms: 15x+21 = 15x + 21
Subtract 15x from both sides: 15x-15x+21 = 15x-15x + 21
21 = 21, which is true for all x. So identity.

Step5: Solve equation 29

12y + 48-4y = 8(y - 6)
Combine like - terms on the left side: 8y + 48 = 8(y - 6)
Distribute the 8 on the right side: 8y + 48 = 8y-48
Subtract 8y from both sides: 8y-8y + 48 = 8y-8y-48
48=-48, which is false. So no solution.

Step6: Solve equation 30

8(z + 6)=4(2z + 12)
Distribute the 8 and 4: 8z+48 = 8z + 48
Subtract 8z from both sides: 8z-8z+48 = 8z-8z + 48
48 = 48, which is true for all z. So identity.

Step7: Solve equation 31

2a + 2 = 3(a + 2)
Distribute the 3: 2a + 2 = 3a+6
Subtract 2a from both sides: 2a-2a + 2 = 3a-2a + 6
2=a + 6
Subtract 6 from both sides: 2-6=a + 6-6
a=-4. So one solution.

Step8: Solve equation 32

\frac{1}{4}x + 5=\frac{1}{4}x
Subtract \frac{1}{4}x from both sides: \frac{1}{4}x-\frac{1}{4}x + 5=\frac{1}{4}x-\frac{1}{4}x
5 = 0, which is false. So no solution.

Step9: Solve equation 33

7(c + 9)=7c + 63
Distribute the 7: 7c+63 = 7c + 63
Subtract 7c from both sides: 7c-7c+63 = 7c-7c + 63
63 = 63, which is true for all c. So identity.

Step10: Solve equation 34

4k + 3=\frac{1}{4}(8k + 16)
Distribute the \frac{1}{4}: 4k + 3 = 2k+4
Subtract 2k from both sides: 4k-2k + 3 = 2k-2k+4
2k + 3 = 4
Subtract 3 from both sides: 2k+3 - 3 = 4 - 3
2k = 1
Divide both sides by 2: k=\frac{1}{2}. So one solution.

Step11: Solve equation 35

3b-13 + 4b = 7b + 1
Combine like - terms on the left side: 7b-13 = 7b + 1
Subtract 7b from both sides: 7b-7b-13 = 7b-7b + 1
-13 = 1, which is false. So no solution.

Step12: Solve equation 36

\frac{1}{2}(\frac{1}{2}m - 8)=\frac{1}{4}(m - 16)
Distribute the \frac{1}{2} on the left side: \frac{1}{4}m-4=\frac{1}{4}(m - 16)
Distribute the \frac{1}{4} on the right side: \frac{1}{4}m-4=\frac{1}{4}m-4
Subtract \frac{1}{4}m from both sides: \frac{1}{4}m-\frac{1}{4}m-4=\frac{1}{4}m-\frac{1}{4}m-4
-4=-4, which is true for all m. So identity.

Answer:

  1. No solution
  2. No solution
  3. Identity
  4. Identity
  5. No solution
  6. Identity
  7. One solution (a = - 4)
  8. No solution
  9. Identity
  10. One solution (k=\frac{1}{2})
  11. No solution
  12. Identity