5=3x+2 3x+2-2=5-2 3x=3 ------ 3 x=1
I understand the equation to be y = -3x - 2 and the point to be (5,1). I substitute 1 for each appearance of y and 5 for each appearance of x: 1 = -3(5) - 2 = -15 - 2 = -17, which is not a true statement. Therefore, that is not a solution. To get a solution, set x=1, and calculate y by substituting this value (1) for x wherever it appears: y = -3(1) - 2 = -3 - 2 = -5. Therefore, (-5,1) is a solution. (I suspect that this is what you meant to put in the question.)
(x + 1)(3x + 10)
3x**2+x-10 Multiply first and last terms 3*-10=-30 Find factors of -30 that add up to second term 6+-5=1 rewrite 3x**2-5x+6x-10 Factor pairs of terms 3x**2-5x+6x-10 = x(3x-5)+2(3x-5) Factor x(3x-5)+2(x-5) = (x+2)(3x-5)
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The solution to 5 = 3x + 2 is found by first subtracting 2:3 = 3x.Then dividing by 3:x = 1.
Yes, it is.
x=1, y=5 is a legitimate answer.
If: y = 3x+4 and y = 2-7x Then: 3x+4 = 2-7x So: 3x+7x = 2-4 => 10x = -2 => x = -1/5 By substitution: y = 17/5 Solution: x = -1/5 and y = 17/5
I understand the equation to be y = -3x - 2 and the point to be (5,1). I substitute 1 for each appearance of y and 5 for each appearance of x: 1 = -3(5) - 2 = -15 - 2 = -17, which is not a true statement. Therefore, that is not a solution. To get a solution, set x=1, and calculate y by substituting this value (1) for x wherever it appears: y = -3(1) - 2 = -3 - 2 = -5. Therefore, (-5,1) is a solution. (I suspect that this is what you meant to put in the question.)
3x+5 Solution- 3x+5=0 3x= (-5) x= (-5)/3
5=3x+2 3x+2-2=5-2 3x=3 ------ 3 x=1
2x+3 = 3x+5 2x-3x = 5-3 -x = 2 x = -2
3x-5=3x+5 There is no solution.
3x+5=2, 3x=2-5 (sign changes) 3x=-3 x=-1
-2
2(x+5) =3X+1 => 2x+10-3x-1=0. => -x+9 => x=9