Answer:
The question is not so clear and complete
Step-by-step explanation:
But for questions like this, since the equation has been given, what is expected is for us to make comparison, compare the RHS with the LHS or by method of comparing coefficients.
We follow the stated conditions since we are told that b and c are both integers which are greater than 1 and b is less than the product of cb. from these conditions, we can compare and get the values of b , c and d.
Another approach is to assume values, make assumptions with the stated conditions, however, our assumptions must be valid and correct if we substitute the assumed values of b, c and d in the equation, it must arrive at the same answer for the RHS. i.e LHS = RHS
Um, I don't know if I actually answered the question, but I think that the second one is cheaper to rent.
Equation 1: y = -2x + 1
Equation 2: y = 2x - 3
Since both equations already have y isolated, we are able to simply set the right side of both equations equal to each other. Since we know that the value of y must be the same, we can do this.
-2x + 1 = 2x - 3
1 = 4x - 3
4 = 4x
x = 1
Then, we need to plug our value of x back into either of the original two equations and solve for y. I will be plugging x back into equation 2 above.
y = 2x - 3
y = 2(1) - 3
y = 2 - 3
y = -1
Hope this helps!! :)
Bring it to the form ax + by = c, where a is positive, and there are no fractions in the equation.
Here, we need to add 2/5x to both sides:
2/5x + y = 0
Then multiply everything by 5 to get rid of the fraction
2x + 5y = 0 <==
