Simplify the following expression: (2-3y)+(6+8y)

vazelinahS 2021-10-20 Answered
Simplify the following expression:
(2-3y)+(6+8y)
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Jayden-James Duffy
Answered 2021-10-21 Author has 91 answers
(23y)+(6+8y)=23y+6+8y
=(2+6)+(3y+8y)
=8+5y
The final answer:
(23y)+(6+8y)=8+5y
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When we solve an equation, do we suppose that it is true and then work backward?
A couple of days ago I was reading Calculus by James Stewart and I read this:
Sometimes it is useful to imagine that your problem is solved and work backward, step by step, until you arrive at the given data. Then you might be able to reverse your steps and thereby construct a solution to the original problem. This procedure is commonly used in solving equations. For instance, in solving the equation 3x−5=7, we suppose that x is a number that satisfies 3x−5=7 and work backward. We add 5 to each side of the equation and then divide each side by 3 to get x=4. Since each of these steps can be reversed, we have solved the problem.
This sounded strange to me! I have always thought that when we solve an equation we don't suppose that the equation is already satisfied. I have always thought that when we solve an equation we use algbraic property of numbers to obtain a simpler equation that is equivalent to the starting equation. And since the equations are equivalent we don't need to suppose that the initial equation is true, because when the last one is true, is true also the first one.
In other words: if I have to solve 3x−5=7 I don't need to suppose that x is a number that satisys 3x−5=7, I simply ad 5 to bothe sides to obtain the equivalent equation 3x=12, then I divide both sides by 3 to obtain the equivalent equation x=4, when the last one is true, is true also the first one and vice versa, the last one is true when x is replaced by 4, so 4 is the solution.
And so this is my question: is it true that when we solve an equation we implicitly suppose that x satisfies the equation (and we need to do that) to apply algebraic properties that give us the equivalent and simplier equation? Do we need this logic assumption in solving equations?
Thanks.
EDIT
Let me try to explain in a better way why I don't understand the need to suppose that x satisfies the equation (i.e. that x makes true the equality). Excuse me for the lenght of this edit.
Let's say I want to solve in R the equation 3x−5=7.
I know that
∀a,b,c∈R,a+c=b+c↔a=b(P1)
and I know that
∀a,b,c∈R,(c≠0→(ac=bc↔a=b))(P2)
Property P1 says to me that:
if a+c=b+c is true, then a=b is true;
if a=b is true, then a+c=b+c is true;
if a+c=b+c is false, then a=b is false;
if a=b is false, then a+c=b+c is false;
and property P2 says to me that, if c≠0, then
if ac=bc is true, then a=b ia true;
if a=b is true, then ac=bc is true;
if ac=bc is false, then a=b is false;
if a=b is false, then ac=bc is false;
If I know all of this, then starting from 3x−5=7 I don't need to suppose that x makes true the equality, because:
I can say that 3x−5=7 is equivalent to 3x=12 because of P1 without supposing that 3x−5=7 is true, they have the same truth value for the same value of x;
from 3x=12, I can say that it is equivalent to x=4 because of P2 without supposing that 3x=12 is true, they have the same truth value for the same value of x;
now I can say that 3x−5=12 is equivalent to x=4 without supposing that 3x−5=12 is true, they have the same truth value for the same value of x;
in the end I have the solution, because when x=4 is true, also 3x−5=7, so 4 is the solution.
What am I doing wrong? Why do I need to suppose that x satisfy the equation?When we solve an equation, do we suppose that it is true and then work backward?
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