5
Y²-5Y+4=0 Y1=-(-5/2) - Square root of ((-5/2)²-4) Y1= 2.5 - Square root of 2.25 Y1 = 1 Y2=-(-5/2) + Square root of ((-5/2)²-4) Y2= 2.5 + Square root of 2.25 Y2 = 4 Y can be either 1 or 4
2/y + 1 + 3 + 5y
Determine the perfect square factors of the term. Then, reduce 'em! √(80x²y) = √(16 * 5 * x² * y) = 4x√(5y)
y2 - 15 = 10Add 15 to each side:y2 = 25Take the square root of each side:y = +5y = -5
X = 17.7 + 0.1y
5y=25x 25x=5y -5y 25x-5y=0
Y²-5Y+4=0 Y1=-(-5/2) - Square root of ((-5/2)²-4) Y1= 2.5 - Square root of 2.25 Y1 = 1 Y2=-(-5/2) + Square root of ((-5/2)²-4) Y2= 2.5 + Square root of 2.25 Y2 = 4 Y can be either 1 or 4
Substitution method: from first equation y = 5x - 8. In the second equation this gives 25x - 5(5x - 8) = 32 ie 25x - 25x + 40 = 32 ie 40 = 32 which is not possible, so the system has no solution. Multiplication method: first equation times 5 gives 25x - 5y = 40, but second equation gives 32 as the value of the identical expression. No solution.
No solution
2/y + 1 + 3 + 5y
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (5 plus or minus the square root of 73) divided by 12. y = 1.128666978776461 y = -0.2953336454431275
To find the y-intercept of the equation 5y = 25x - 15, we first need to isolate y. Dividing both sides by 5 gives us y = 5x - 3. The y-intercept occurs when x = 0, so substituting x = 0 into the equation gives us y = -3. Therefore, the y-intercept of the equation is -3.
Determine the perfect square factors of the term. Then, reduce 'em! √(80x²y) = √(16 * 5 * x² * y) = 4x√(5y)
y2 - 15 = 10Add 15 to each side:y2 = 25Take the square root of each side:y = +5y = -5
3x/7 + 5y/14x = 3x*2x/(7*2x) + 5y/14x = (6x2 + 5y)/14x
X = 17.7 + 0.1y
Area of square: 25y^2