Certainly! Let's check if the values given in the brackets are indeed solutions to their respective equations.
### (a) [tex]\(2x + 4 = 15\)[/tex]
Given value: [tex]\(x = 2\)[/tex].
To check, substitute [tex]\(x\)[/tex] with 2:
[tex]\[2(2) + 4 = 4 + 4 = 8\][/tex]
However, the right-hand side (RHS) is 15. Clearly, [tex]\(8 \neq 15\)[/tex].
Thus, [tex]\(x = 2\)[/tex] is not a solution.
### (b) [tex]\(7x + 15 = 45\)[/tex]
Given value: [tex]\(x = 5\)[/tex].
To check, substitute [tex]\(x\)[/tex] with 5:
[tex]\[7(5) + 15 = 35 + 15 = 50\][/tex]
However, the RHS is 45. Clearly, [tex]\(50 \neq 45\)[/tex].
Thus, [tex]\(x = 5\)[/tex] is not a solution.
### (c) [tex]\(7x + 2 = 23\)[/tex]
Given value: [tex]\(x = 3\)[/tex].
To check, substitute [tex]\(x\)[/tex] with 3:
[tex]\[7(3) + 2 = 21 + 2 = 23\][/tex]
The RHS is also 23. Clearly, [tex]\(23 = 23\)[/tex].
Thus, [tex]\(x = 3\)[/tex] is a solution.
### (d) [tex]\(\frac{4x}{5} + 2 = 6\)[/tex]
Given value: [tex]\(x = 5\)[/tex].
To check, substitute [tex]\(x\)[/tex] with 5:
[tex]\[\frac{4(5)}{5} + 2 = \frac{20}{5} + 2 = 4 + 2 = 6\][/tex]
The RHS is also 6. Clearly, [tex]\(6 = 6\)[/tex].
Thus, [tex]\(x = 5\)[/tex] is a solution.
### (e) [tex]\(4p - 5 = 16\)[/tex]
Given value: [tex]\(p = 7\)[/tex].
To check, substitute [tex]\(p\)[/tex] with 7:
[tex]\[4(7) - 5 = 28 - 5 = 23\][/tex]
However, the RHS is 16. Clearly, [tex]\(23 \neq 16\)[/tex].
Thus, [tex]\(p = 7\)[/tex] is not a solution.
### (f) [tex]\(4p - 5 = 23\)[/tex]
Given value: [tex]\(p = 7\)[/tex].
To check, substitute [tex]\(p\)[/tex] with 7:
[tex]\[4(7) - 5 = 28 - 5 = 23\][/tex]
The RHS is also 23. Clearly, [tex]\(23 = 23\)[/tex].
Thus, [tex]\(p = 7\)[/tex] is a solution.
### (g) [tex]\(2x + 5 = 15\)[/tex]
Given value: [tex]\(x = 5\)[/tex].
To check, substitute [tex]\(x\)[/tex] with 5:
[tex]\[2(5) + 5 = 10 + 5 = 15\][/tex]
The RHS is also 15. Clearly, [tex]\(15 = 15\)[/tex].
Thus, [tex]\(x = 5\)[/tex] is a solution.
### (h) [tex]\(3x - 4 = 16\)[/tex]
Given value: [tex]\(x = 2\)[/tex].
To check, substitute [tex]\(x\)[/tex] with 2:
[tex]\[3(2) - 4 = 6 - 4 = 2\][/tex]
However, the RHS is 16. Clearly, [tex]\(2 \neq 16\)[/tex].
Thus, [tex]\(x = 2\)[/tex] is not a solution.
### (i) [tex]\(\frac{2x}{5} + 4 = 10\)[/tex]
Given value: [tex]\(x = 5\)[/tex].
To check, substitute [tex]\(x\)[/tex] with 5:
[tex]\[\frac{2(5)}{5} + 4 = \frac{10}{5} + 4 = 2 + 4 = 6\][/tex]
However, the RHS is 10. Clearly, [tex]\(6 \neq 10\)[/tex].
Thus, [tex]\(x = 5\)[/tex] is not a solution.
Summarizing:
a) False, b) False, c) True, d) True, e) False, f) True, g) True, h) False, i) False
