Solve for x:x/5 - 2 = x/2 + 3
Put each term in x/5 - 2 over the common denominator 5: x/5 - 2 = x/5 - (10)/5:x/5 - (10)/5 = x/2 + 3
x/5 - (10)/5 = (x - 10)/5:(x - 10)/5 = x/2 + 3
Put each term in x/2 + 3 over the common denominator 2: x/2 + 3 = x/2 + 6/2:(x - 10)/5 = x/2 + 6/2
x/2 + 6/2 = (x + 6)/2:(x - 10)/5 = (x + 6)/2
Multiply both sides by 10:(10 (x - 10))/5 = (10 (x + 6))/2
10/5 = (5×2)/5 = 2:2 (x - 10) = (10 (x + 6))/2
10/2 = (2×5)/2 = 5:2 (x - 10) = 5 (x + 6)
Expand out terms of the left hand side:2 x - 20 = 5 (x + 6)
Expand out terms of the right hand side:2 x - 20 = 5 x + 30
Subtract 5 x from both sides:(2 x - 5 x) - 20 = (5 x - 5 x) + 30
2 x - 5 x = -3 x:-3 x - 20 = (5 x - 5 x) + 30
5 x - 5 x = 0:-3 x - 20 = 30
Add 20 to both sides:(20 - 20) - 3 x = 20 + 30
20 - 20 = 0:-3 x = 30 + 20
30 + 20 = 50:-3 x = 50
Divide both sides of -3 x = 50 by -3:(-3 x)/(-3) = 50/(-3)
(-3)/(-3) = 1:x = 50/(-3)
Multiply numerator and denominator of 50/(-3) by -1:Answer: x = (-50)/3
It depends on what variable you are tying to solve for first. Say you are trying to solve for x first and then y on the first problem you wrote.
In substitution you solve one of the equations for example with
6x+2y=-10
2x+2y=-10
you solve 2x+2y=-10 for x
2x+2y=-10
-2y = -2y (what you do to one side of the = you do to the other)
2x=-10-2y (to get the variable by its self you divide the # and the variable)
/2=/2 (-10/2=-5 and -2y/2= -y or -1y, they are the same either way)
x=-5-y
now you put that in your original equation that you didn't solve for:
6(-5-y)+2y=-10 solve for that
-30-6y+2y=-10 combine like terms
-30-4y=-10 get the y alone and to do this you first get the -30 away from it
+30=+30
-4y=20 divide the -4 from each side
/-4=/-4 (20/-4=-5)
y=-5
now the equation you previously solved for x can be solved for y.
x=-5-y
x=-5-(-5) a minus parenthesis negative -(- gives you a positive
-5+5=0
x=0
and now we have solved the problem. x=0 and y=-5
If in the triangle ABC , BF is an angle bisector and ∠ABF=41° then angle m∠BCE=8°.
Given that m∠ABF=41° and BF is an angle bisector.
We are required to find the angle m∠BCE if BF is an angle bisector.
Angle bisector basically divides an angle into two parts.
If BF is an angle bisector then ∠ABF=∠FBC by assuming that the angle is divided into two parts.
In this way ∠ABC=2*∠ABF
∠ABC=2*41
=82°
In ΔECB we got that ∠CEB=90° and ∠ABC=82° and we have to find ∠BCE.
∠BCE+∠CEB+EBC=180 (Sum of all the angles in a triangle is 180°)
∠BCE+90+82=180
∠BCE=180-172
∠BCE=8°
Hence if BF is an angle bisector then angle m∠BCE=8°.
Learn more about angles at brainly.com/question/25716982
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A and B Definitely. 3 and 17, -3 and -17
