Which equation is an identity? -
() 7-(9x+3)=-9x-4
() 6m-5=7m+5-m
() 10p+6-p=12p-3(p-2)
() 3y+2=3y-2

Which equation has no solution? -
() 7v+2=8v-3
() 3x-5=3x+8-x
() 4y+5=4y-6
() 7z+6=-7z-5

Solve the equation.
5+7x=11+7x -
() 0
() 14
() infinitely solutions
() no solution

Answer:

Question 1). Option C.

Question 2) Option C.

Question 3) Option D.

Step-by-step explanation:

Question 1), A. 7 - (9x + 3) = -9x -4

7 - 9x - 3 = -9x - 4

-9x + 4 = -9x - 4

Left hand side(L.H.S.)≠ Right hand side(R.H.S.)

Therefore, it's not an identity

B). 6m - 7 = 7m + 5 -m

6m -7 = 6m + 5

Again L.H.S.≠R.H.S.

So, it's not an identity.

C). 10p + 6 - p = 12p - 3(p - 2)

9p + 6 = 12p - 3p + 6

9p + 6 = 9p + 6

L.H.S.=R.H.S.

Therefore, it's an identity.

D). 3y + 2 = 3y - 2

L.H.S. ≠ R.H.S.

Therefore, it's not an identity.

Question 2. Part A. 7v + 2 = 8v - 3

7v - 8v = -2 - 3

- v = - 5

v = 5

Part B. 3x - 5 = 3x + 8 - x

3x - 5 = 2x + 8

3x - 2x = 8 + 5

x = 13

Part C. 4y + 5 = 4y - 6

This equation has no solution.

Part D. 7z + 6 = -7z - 5

7z + 7z = -6 - 5

14z = -11

z = [tex]-\frac{11}{14}[/tex]

Question 3). 5 + 7x = 11 + 7x

This equation has same coefficient of variable x on both the sides of the equation.

Therefore, equation has no solution.

Option D. no solution is the correct option.

The correct option for different parts are as follows:

Part (1): [tex]\boxed{\bf option (c)}[/tex]

Part (2): [tex]\boxed{\bf option (c)}[/tex]

Part (3): [tex]\boxed{\bf option (d)}[/tex]

Further explanation:

Part (1):

Option (a)

Here, the equation is [tex]7-(9x+3)=-9x-4[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}7-(9x+3)&\ _{=}^{?}-9x-4\\7-9x-3&\ _{=}^{?}-9x-4\\4-9x&\neq-9x-4\end{aligned}[/tex]  

Here, left hand side (LHS) is not equal to right hand side (RHS).

Therefore, the given equation is not an identity.

This implies that option (a) is incorrect.

Option (b)

Here, the equation is [tex]6m-5=7m+5-m[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}6m-5\ &_{=}^{?}\ 7m+5-m\\6m-5 &\neq6m+5\end{aligned}[/tex]  

Here, left hand side (LHS) is not equal to right hand side (RHS).

Therefore, the given equation is not the identity.

This implies that option (b) is incorrect.

Option (c)

Here, the equation is [tex]10p+6-p=12p-3(p-2)[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}10p+6-p\ &_{=}^{?}\ 12p-3(p-2)\\9p+6\ &_{=}^{?}\ 12p-3p+6\\9p+6&\neq9p+6\end{aligned}[/tex]  

Here, left hand side (LHS) is equal to right hand side (RHS).

Therefore, the given equation is an identity.

This implies that option (c) is correct.

Option (d)

Here, the equation is [tex]3y+2=3y-2[/tex].

Now, the above equation is as follows:

[tex]3y+2\neq3y-2[/tex]  

Here, left hand side (LHS) is not equal to right hand side (RHS).

Therefore, the given equation is not an identity.

This implies that option (d) is incorrect.

Therefore, equation in option (c) is an identity.  

Part (2):

Option (a)

Here, the equation is [tex]7v+2=8v-3[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}7v+2&=8v-3\\7v-8v&=-2-3\\-v&=-5\\v&=5\end{aligned}[/tex]  

Thus, the value of [tex]v[/tex]is [tex]5[/tex].

Therefore, the given equation has a solution.

This implies that option (a) is incorrect.

Option (b)

Here, the equation is [tex]3x-5=3x+8-x[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}3x-5&=3x+8-x\\3x-3x+x&=8+5\\x&=13\end{aligned}[/tex]  

Thus, the value of [tex]x[/tex] is [tex]5[/tex].

Therefore, the given equation has a solution.

This implies that option (b) is incorrect.

Option (c)

Here, the equation is [tex]4y+5=4y-6[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}4y+5&=4y-6\\4y-4y&=-6-5\\0&\neq-11\end{aligned}[/tex]

Thus, the given equation has no solution.

This implies that option (c) is correct.

Option (d)

Here, the equation is [tex]7z+6=-7z-5[/tex].

Now, solve the above equation as follows:

[tex]\begin{aligned}7z+6&=-7z-5\\7z+7z&=-5-6\\14z&=-11\\z&=-\dfrac{11}{14}\end{aligned}[/tex]

Thus, the value of [tex]z[/tex] is [tex]-\frac{11}{14}[/tex].

Therefore, the given equation has a solution.

This implies that option (d) is incorrect.

Therefore, equation in option (c) does not have solution.

Part (3):

The equation is [tex]5+7x=11+7x[/tex].

Solve the above equation as follows:

[tex]\begin{aligned}5+7x&=11+7x\\7x-7x&=11-5\\0&\neq6\end{aligned}[/tex]

Therefore, the given equation has no solution.

Option (a)

Here, the value of [tex]x[/tex] is [tex]0[/tex].

But the equation [tex]5+7x=11+7x[/tex] has no solution.

So, option (a) is incorrect.

Option (b)

Here, the value of [tex]x[/tex] is [tex]14[/tex].

But the equation [tex]5+7x=11+7x[/tex] has no solution.

So, option (b) is incorrect.

Option (c)

In option (c) it is given that there are infinite number of solutions.

But the equation [tex]5+7x=11+7x[/tex] has no solution.

So, option (c) is incorrect.

Option (d)

In option (d) it is given that the solution does not exist.

As per our calculation the equation [tex]5+7x=11+7x[/tex] does not have any solution.

So, option (d) is correct.

Learn more

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Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Linear equations

Keywords: Linear equations, linear equation in one variable, linear equation in two variable, slope of a line, equation of the line, function, real numbers, ordinates, abscissa, interval, open interval, closed intervals, semi-closed intervals, semi-open intervals, sets, range domain, codomain.

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