answer:To solve 10 - 3(x + 2) = 4: Begin by distributing the -3 to the values inside the (): 10 - 3x - 6 = 4 Combine like terms on the left side of the equation (10 - 6): 4 - 3x = 4 Isolate the term with the variable by additive inverse 4 cancel(-4) - 3x = 4 - 4 -3x = 0 Isolate the variable by multiplicative inverse (cancel(-3)x)/cancel(-3) = 0/-3 x = 0 Therefore, the value of x is 0.

answer:The real solution to the equation is given by: x = frac{1}{722} left(-29077+54 sqrt{226534}right) Note that this solution assumes the domain restriction where both square roots are real, which implies that -frac{29 x}{2}-frac{5}{2} geq 0 and 6-5 x geq 0.

answer:The given equation represents a situation where we need to find the value of x that satisfies the equality between the sum of two square roots and a constant. To solve for x, it's best to square both sides of the equation to eliminate the square roots, then isolate x: Starting with: [ sqrt{10 - frac{5x}{2}} + sqrt{frac{25}{2} - frac{x}{2}} = frac{29}{2} ] Square both sides: [ left(sqrt{10 - frac{5x}{2}} + sqrt{frac{25}{2} - frac{x}{2}}right)^2 = left(frac{29}{2}right)^2 ] Expand and simplify: [ 10 - frac{5x}{2} + 2sqrt{left(10 - frac{5x}{2}right)left(frac{25}{2} - frac{x}{2}right)} + frac{25}{2} - frac{x}{2} = frac{841}{4} ] Combine like terms: [ 20 - 5x + 25 - x + 2sqrt{50 - 15x - frac{5x^2}{4}} = frac{841}{4} ] Simplify further: [ 45 - 6x + 2sqrt{-frac{5x^2}{4} - 15x + 50} = frac{841}{4} ] Multiply both sides by 4 to get rid of fractions: [ 180 - 24x + 8sqrt{-5x^2 - 60x + 200} = 841 ] Subtract 180 from both sides: [ -24x + 8sqrt{-5x^2 - 60x + 200} = 661 ] Divide by 8: [ -3x + sqrt{-5x^2 - 60x + 200} = frac{661}{8} ] Square both sides again to get rid of the square root: [ (-3x)^2 - 2(-3x)sqrt{-5x^2 - 60x + 200} + (-5x^2 - 60x + 200) = left(frac{661}{8}right)^2 ] Expand and simplify: [ 9x^2 + 30xsqrt{-5x^2 - 60x + 200} - 5x^2 - 60x + 200 = frac{43681}{64} ] Combine like terms: [ 4x^2 + 30xsqrt{-5x^2 - 60x + 200} - 60x - frac{43681}{64} + 200 = 0 ] This is a complicated equation, but we can recognize that the expression under the square root must be zero for the equation to have a real solution. Thus, [ -5x^2 - 60x + 200 = 0 ] Divide by -5: [ x^2 + 12x - 40 = 0 ] Factor the quadratic equation: [ (x + 16)(x - 2) = 0 ] Solving for x gives us two possible solutions: [ x = -16 quad text{or} quad x = 2 ] However, we need to check these solutions in the original equation, as squaring both sides can introduce extraneous solutions. Upon substitution, only x = 2 satisfies the original equation. Therefore, the only real solution is x = 2. Final Answer: x = 2

answer:The next iterate of the Householder method is given by: a_{(6)}=b+frac{b left(b^{16}+12 b^{12}+36 b^8+20 b^4+1right)}{b^3 left(b^{16}+15 b^{12}+60 b^8+60 b^4+15right)}