Equivalent equations are simply equations that have the same value, irrespective of their form
1. 7 + 5x = 3 - 2x and 7 + 7x = 3
Start by solving for x in both equations
[tex]\mathbf{7 + 5x = 3 - 2x}[/tex]
Collect like terms
[tex]\mathbf{2x + 5x = 3 - 7}[/tex]
[tex]\mathbf{7x = - 4}[/tex]
Divide both sides by 7
[tex]\mathbf{x = -\frac 47}[/tex]
[tex]\mathbf{7 + 7x = 3}[/tex]
Collect like terms
[tex]\mathbf{7x = 3-7}[/tex]
[tex]\mathbf{7x = - 4}[/tex]
Divide both sides by 7
[tex]\mathbf{x = -\frac 47}[/tex]
The values of x in both equations are the same;
Hence, 7 + 5x = 3 - 2x and 7 + 7x = 3 are equivalent
2. 3 (x – 4) = 15 and x-4 = 15
Rewrite the first equation as follows
[tex]\mathbf{3 (x - 4) = 15 }[/tex]
Divide both sides by 3
[tex]\mathbf{x - 4 = 5 }[/tex]
In the second equation, we have:
[tex]\mathbf{x-4 = 15}[/tex]
The values of x - 4 are not the same in both equations.
Hence, 3(x – 4) = 15 and x-4 = 15 are not equivalent
3. x2 = 6x and x = 6
Solve for x in the first equation as follows:
[tex]\mathbf{x^2 = 6x}[/tex]
Divide both sides by x
[tex]\mathbf{x = 6}[/tex]
The results of both equations are the same.
Hence, [tex]\mathbf{x^2 = 6x}[/tex] and [tex]\mathbf{x = 6}[/tex] are equivalent
4. (x+5) = 10 and 1 = 10(x – 5)
Solve for x in the first equation as follows
[tex]\mathbf{x+5 = 10}[/tex]
[tex]\mathbf{x = 5}[/tex]
Solve for x in the second equation as follows
[tex]\mathbf{1 = 10(x - 5)}[/tex]
Divide both sides by 10
[tex]\mathbf{0.1 =x - 5}[/tex]
Add 5 to both sides
[tex]\mathbf{x = 5.1}[/tex]
The values of [tex]\mathbf{x+5 = 10}[/tex] and [tex]\mathbf{1 = 10(x - 5)}[/tex] are not the same
Hence, [tex]\mathbf{x+5 = 10}[/tex] and [tex]\mathbf{1 = 10(x - 5)}[/tex] are not equivalent
Read more about equivalent equations at:
https://brainly.com/question/2972832
