a = 4
-2a + 5 = 1 -2a = -5 + 1 -2a = -4 a = -4/-2 a = 2
a = 1
(2a+1)(a+1)
2a-6a-a=b+4b-4b -5a=5b-4b -5a=1b a=-5 b=1
a2 + 2a - 15 = 0This can be factorised:a2 + 2a - 15 = (a - 3)(a + 5) = 0So a = 3 and -5.A more robust way is to shove it into the general solution:If αx2 + βx + η = 0, then x = (-β ± √(β2 - 4αγ)) / 2αSo if a2 + 2a - 15 = 0, then:a = (-2 ± √(22 - 4*1*-15)) / 2*1a = (-2±√(64))/2 = (-2±8)/2a = -10/2, 6/2a = -5, 3
1
-2a + 5 = 1 -2a = -5 + 1 -2a = -4 a = -4/-2 a = 2
a = 1
2a + 1 = 3subtract 1 from both sides2a = 2divide both sides by 2a = 1
If ba = 2a + b , then find 23 + 32 +1
If: 2a+13 = 5a+1 Then: a = 4
a+a=1 2a=1 a=1/2
Do the problem like this: 2a - 15 = -1 Add 15 to both sides: 2a = (-1) + 15 Simplify 2a = 14 Divide by 2 to get the variable by itself: a = 7 Now, check the problem: 2(7) - 15 = -1 14 - 15 = -1 This is true, so a = 7 is a good solution.
6a+1=2a+25 Subtract 2a from both sides 4a+1=25 Subtract 1 from both sides 4a=24 Divide both sides by 4 a=6 That is your final answer.
(2a+1)(a+1)
2a-6a-a=b+4b-4b -5a=5b-4b -5a=1b a=-5 b=1
a2 + 2a - 15 = 0This can be factorised:a2 + 2a - 15 = (a - 3)(a + 5) = 0So a = 3 and -5.A more robust way is to shove it into the general solution:If αx2 + βx + η = 0, then x = (-β ± √(β2 - 4αγ)) / 2αSo if a2 + 2a - 15 = 0, then:a = (-2 ± √(22 - 4*1*-15)) / 2*1a = (-2±√(64))/2 = (-2±8)/2a = -10/2, 6/2a = -5, 3