a) 3(x+1)+x+2=2(2x+1)+3
I distributed the equation and got 5=5 as my final answer.
b) DON'T UNDERSTAND
is the resulting equation from part (a) always true,sometimes ture, or never true? Explain your reasoning.
please help me
If you solve an equation and your variables drop out like in the one above, two cases are possible
1. the resulting statement is true , in which case your equation was an "identity" , and it is true for all values of the variable
or
2. the resulting statement is false, in which case there is no solution at all to your equation
e.g. If your equation had been
3(x+1)+x+2=2(2x+1)+2
you would end up with
5 = 4, a false statement.
Thus your equation would have no solution.
So for your given equation, the statement is always true.
To solve the given equation: 3(x+1)+x+2=2(2x+1)+3
a) Distribute on both sides of the equation:
3x + 3 + x + 2 = 4x + 2 + 3
This simplifies to:
4x + 5 = 4x + 5
b) Now, let's analyze the resulting equation: 4x + 5 = 4x + 5
The equation implies that both sides are equal. Here, you may observe that no matter what value of 'x' you substitute, the equation will always hold true.
So, the resulting equation from part (a) is always true. This means that any value of 'x' you substitute will satisfy the equation. In other words, the equation has an infinite number of solutions.