Seven less than the quotient of x and 9
1 answer:
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Answer:
The fifth line contains the error.
Step-by-step explanation:
Line 1 is the equation.
Lines 2, 3, and 4 are correct.
The error is in the 5th line.
The square root of a negative number is not a real number, so if this problem is solved only with real numbers, there is no solution.
If imaginary numbers are allowed, then the last line should read:
Answer:
Step-by-step explanation:
for two lines to be parallel they MUST have the same slope but have different Y intercepts ( Y axis crossing values)
parallel lines will never intersect each other
first I use a graphing calcuator to solve the problem
I like to use y intercept form better, but you don't have use it
I will try to solve it directly without going through the y intercept equation
x + 5y = 10 in y intercept form y = mx + b is
5y = -x + 10 divide both sides by 5
y = -x/5 + 10/5
y = -x/5 + 2
Now I need to find a parallel line that passes through the x =1 and y = 3 point recall this line will have the same slope = m = -1/5
and a different Y axis crossing point 'b' that we don't know
y = -x/5 + b y = 3 when x = 1 so slove for 'b'
3 = -1/5 + b
3 = -0.2 + b add -0.2 to both sides
3.2 = b
y = -x/5 + 3.2 this is answer in y intercept form
if you multiply both sides by 5
5y = -x + 16 add x to both sides
x + 5y = 16 answer in standard form
THE EASIER WAY TO SOLVE THE PROBLEM IS ......
x + 5y = 10 find a line parallel that passes through x = 1, y = 3
x + 5y = Z Z is an unknown value plug in x and y and solve for Z
1 + 5(3) = Z
Z = 16 so the parallel line in standard form is
x + 5y = 16
same as before, my y intercept method was correct, but not worth the effort
