0
By adding one to each side of the equation. The general approach to this kind of problems is to "isolate" the variable (in this case, "x") on one side of the equation, getting rid of anything else. However, you must always do the same thing on both sides of the equation, otherwise, you will get wrong results.
Add your answer:
Earn +20 pts
Maxz equals 2x1 plus 2x2 stc 5x1 plus 3x2 equals 8 x1 plus 2x2 equals 4 x1 x2 equals 0 and integerssolve by ipp?
3
What property is shown in 4 x1 equals 4?
The property that 1 is the identity for multiplication.
How do i solve this equation . C-2 - 4 equals 3?
You would do it it by oppisites. 3+2+4=9 which equals C.
How do you solve this equation -a equals -4?
Easy! if -a = -4 then a = 4!!
What would you do to solve 2y plus 4 equals 3y?
2y + 4 = 3ySubtract 2y from both sides:4 = ywhew
What times 4 equals 3?
To solve the equation "What times 4 equals 3," you would need to divide 3 by 4. The answer is 0.75. This is because 4 multiplied by 0.75 equals 3.
If y equals -2 when x equals 4 find x when y equals 5?
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
How do you solve y-4 equals x?
y=x+4
12 equals 3 x blank equals 6 x blank equals blank x 1 fill in the blanks?
12 equals 3x(4) equals 6x(2) equals (12)x1
M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
How do you solve 3p equals 4?
you cant unless you know what the value of P equals
9x-5y-4 equals 0 Solve for y?
Can't be done unless you have another equation with the same x and y. Then you would solve for simultaneous equations.
