x > (15-2x)/-3
looks like you multiplied both sides by -3, but did not reverse the inequality sign.
should have been:
-3x < 15 - 2x
-x < 15
x> -15
{x | x > -15 }
2. I would use substitution.
from the first: y =9-x
sub into the 2nd:
3(9-x) = 2x+8
27 - 3x = 2x + 8
-5x = -19
x = 19/5 = 3.8
y = 9-3.8 = 5.2
1. x>15-2x / -3 Solve the linear inequality. Write the solution in set builder notation.
So, -3x>15-2x, x<-15..?
And how do I write it in set builder notation?
2. y+x=9 and 3y=2x + 8. How do I solve this system of equations?
3. Find the maximum y- value of the following quadratic function: f(x)= -2x^2 + 7x + 1
4. Solve the following equation: 4x/3 + 1/2 = 5y/6
I'm not even sure what this question is asking...
5. Find the solution set to the equation below: 2*absolute value of x-100 = 50 +absolute value of x-100.
x = 150?
2 answers
3. Find the maximum y- value of the following quadratic function: f(x)= -2x^2 + 7x + 1
The maximum value of the y value is obtained from the vertex
the x of the vertex is -b/(2a) = -7/-4 = 7/4
y = -2(49/16) + 7(7/4) + 1 = 57/8
4.
4x/3 + 1/2 = 5y/6
multiply each term by 6
8x + 3 = 5y
8x + 5y + 3 = 0
----> the equation of a straight line, with slope of -5/3
"solve" is not the correct instruction for this equation.
5.
2|x-100| = 50 + |x-100|
2(x-100) = 50 + |x-100| OR -2(x-100) = 50 + |x-100|
case1 :
2(x-100) = 50 + |x-100|
2x - 200 = 50 + |x-100|
2x - 250 = |x-100|
x-100 = 2x-250 OR x-100 = -2x + 250
-x = -150 OR 3x = 350
x = 150 OR x = 350/3
case2:
-2(x-100) = 50 + |x-100|
-2x + 100-50 = |x-100|
-2x + 50 = |x-100|
x-100 = -2x + 50 OR x-100 = 2x - 50
3x = 150 OR -x = 50
x = 50 or x = -50
x = ±50 , 150 , 350/3
check my arithmetic on the last one
The maximum value of the y value is obtained from the vertex
the x of the vertex is -b/(2a) = -7/-4 = 7/4
y = -2(49/16) + 7(7/4) + 1 = 57/8
4.
4x/3 + 1/2 = 5y/6
multiply each term by 6
8x + 3 = 5y
8x + 5y + 3 = 0
----> the equation of a straight line, with slope of -5/3
"solve" is not the correct instruction for this equation.
5.
2|x-100| = 50 + |x-100|
2(x-100) = 50 + |x-100| OR -2(x-100) = 50 + |x-100|
case1 :
2(x-100) = 50 + |x-100|
2x - 200 = 50 + |x-100|
2x - 250 = |x-100|
x-100 = 2x-250 OR x-100 = -2x + 250
-x = -150 OR 3x = 350
x = 150 OR x = 350/3
case2:
-2(x-100) = 50 + |x-100|
-2x + 100-50 = |x-100|
-2x + 50 = |x-100|
x-100 = -2x + 50 OR x-100 = 2x - 50
3x = 150 OR -x = 50
x = 50 or x = -50
x = ±50 , 150 , 350/3
check my arithmetic on the last one