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Q: Ax plus b equals bx plus c then x equals?

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x = -c/(a+b), provided a+b is not 0

x2+bx+ax+ab = x2+ax+bx+ab = x(x+a)+b(x+a) = (x+a)(x+b)

3a+ax+3b+bx = 3(a+b)+(a+b)x = (a+b)(3+x)

Two: one is 0, the other is -b/a ax2 + bx + c = 0, but c = 0 ⇒ ax2 + bx + 0 = 0 ⇒ ax2 + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or (ax + b) = 0 ⇒ x = -b/a

2x - 13x + 42 = x +ax + b a + b = 2(x - 6.5x + 21) = 34 = a + b

Related questions

x = -c/(a+b), provided a+b is not 0

x2+bx+ax+ab = x2+ax+bx+ab = x(x+a)+b(x+a) = (x+a)(x+b)

3a+ax+3b+bx = 3(a+b)+(a+b)x = (a+b)(3+x)

a = 0, b = 0.

Your two equations are: AX + BY = A - B BX - AY = A + B + B Because you have four variables (A, B, X, Y), you cannot solve for numerical values for X and Y. There are a total of four answers to this question, solving each equation for X and Y independently. First equation: X = (A - B - BY)/A Y= (A - B - AX)/B Second equation: X = (A +2B +AY)/B Y = (BX - A - 2B)/A

If x is a null matrix then Ax = Bx for any matrices A and B including when A not equal to B. So the proposition in the question is false and therefore cannot be proven.

(x + 3)(a + b)

6a square plus b square

Two: one is 0, the other is -b/a ax2 + bx + c = 0, but c = 0 ⇒ ax2 + bx + 0 = 0 ⇒ ax2 + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or (ax + b) = 0 ⇒ x = -b/a

Ax + Bx + C is not a trinomial!

(x + 2)(a + b)

double-replacement

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