10(x-2) = 10
one solution
no solution
infinate solutions
-2(y-3)= 2y- 6
one solution
no solution
infinate solutions
I got it, first one was no solutions, second one was one solution.
each has a single solution
(and that's infin>b>ite, not infinate)
To determine the number of solutions for each equation, we need to simplify the equations and evaluate them.
For the first equation, 10(x-2) = 10, we can start by distributing the 10 to the expression inside the parentheses:
10 * x - 10 * 2 = 10.
This results in:
10x - 20 = 10.
Next, we can isolate the variable by moving the constant term (-20) to the other side of the equation:
10x = 10 + 20,
which simplifies to:
10x = 30.
Finally, we can solve for x by dividing both sides of the equation by 10:
x = 30 / 10,
which gives us:
x = 3.
Since we obtained a unique value for x, the equation has one solution.
Now let's move on to the second equation, -2(y-3) = 2y - 6.
Just like before, we'll begin by distributing the -2 to the expression within the parentheses:
-2 * y + 2 * 3 = 2y - 6.
Simplifying further, we have:
-2y + 6 = 2y - 6.
Next, we can gather the y terms on one side of the equation and the constant terms on the other side:
-2y - 2y = -6 - 6 + 6,
which simplifies to:
-4y = -6.
Dividing both sides of the equation by -4, we get:
y = -6 / -4,
which simplifies to:
y = 3/2 or 1.5.
Since we found a unique value for y, the equation also has one solution.
In summary:
For the equation 10(x-2) = 10, there is one solution: x = 3.
For the equation -2(y-3) = 2y - 6, there is one solution: y = 1.5 (or 3/2).
Both equations have a single solution.