Non - Linear Equations

If we put $x = 6a, y = 6b, z = 6c$ , and subtract the second equation from the first, and the third from the second, yields $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}+ \sqrt{c}) \; = \; 6 \; = \; (\sqrt{b}-\sqrt{c})(\sqrt{a}+\sqrt{b}+ \sqrt{c})$ so that $\sqrt{a}-\sqrt{b} = \sqrt{b}-\sqrt{c}$ . Substituting this into the second equation gives $1 \; = \; b - \sqrt{ac} \; = \; \tfrac14(\sqrt{a}+\sqrt{c})^2 - \sqrt{ac} \; = \; \tfrac14(\sqrt{a}-\sqrt{c})^2$ and hence $\sqrt{a} = \sqrt{c} \pm 2$ . Thus $7 \; = \; a - \sqrt{bc} \; = \; (\sqrt{c} \pm 2)^2 - \tfrac12(\sqrt{a}+\sqrt{c})\sqrt{c} \; = \; (\sqrt{c} \pm 2)^2 - \tfrac12(2\sqrt{c} \pm 2)\sqrt{c} \; = \; 4 \pm 3\sqrt{c}$ and so $\sqrt{c} = \pm 1$ . Thus we must take the $+$ option out of $\pm$ , and obtain $\sqrt{a}=3,\sqrt{b}=2,\sqrt{c}=1$ , and hence $x=54, y=24,c=6$ , making the answer $54+24+6 = \boxed{84}$ .

let $a^{2} = x, b^{2} = y, c^{2} = z$

• $a^{2} - bc = 42$ (1)
• $b^{2} - ca = 6$ (2)
• $c^{2} - ab = -30$ (3)

from (1)

• $a^{2} = 42+bc$

solving $a$ from (2)

• $a = \dfrac{b^{2} - 6}{c}$ (4)

solving $a$ from (3)

• $a = \dfrac{c^{2} + 30}{b}$ (5)

multiplying (4) and (5) and equating it to (1)

=> $a^{2} = \dfrac{(b^{2} - 6)(c^{2} + 30)}{bc} = 42+bc$

=> $(bc)^{2} - 6c^{2} + 30b^{2} - 180 = 42bc + (bc)^{2}$

=> $5b^{2} - c^{2} - 30 = 7bc$

=> $5y - z = 30 + 7\sqrt{yz}$

since $\sqrt{yz} = x - 42$

=> $5y-z = 30 + 7(x-42)$ (yey, eq. 1)

from (2)

• $b^{2} = 6 + ca$

solving $b$ from (1)

• $b = \dfrac{a^{2} - 42}{c}$ (6)

solving $b$ from (3)

• $b = \dfrac{c^{2} + 30}{a}$ (7)

multiplying (6) and (7) and equating it to (2)

=> $b^{2} = \dfrac{(a^{2} - 42)(c^{2} + 30)}{ca} = 6+ca$

=> $(ac)^{2} - 42c^{2} + 30a^{2} - 1260 = 6ca + (ca)^{2}$

=> $5a^{2} - 7c^{2} - 210 = ca$

=> $5x - 7z = 210 + \sqrt{zx}$

since $\sqrt{zx} = y - 6$

=> $5x - 7z = 210 + (y-6)$ (yey, eq. 2)

from (3)

• $c^{2} = ab - 30$

solving $c$ from (1)

• $c = \dfrac{a^{2} - 42}{b}$ (8)

solving $c$ from (2)

• $c = \dfrac{b^{2} - 6}{a}$ (9)

multiplying (8) and (9) and equating it to (3)

=> $c^{2} = \dfrac{(a^{2} - 42)(b^{2} - 6)}{ab} = ab - 30$

=> $(ab)^{2} - 42b^{2} - 6a^{2} + 252 = (ab)^{2} - 30ab$

=> $a^{2} + 7b^{2} - 42 = 5ab$

=> $x + 7y = 42 + 5\sqrt{xy}$

since $\sqrt{xy} = z+30$

=> $x + 7y = 42 + 5(z+30)$ (yey, eq. 3)

Equating the 3 (yey) equations, it will give:

$(x,y,z) = (54, 24, 6)$

and hence, $x+y+z = 54+24+6 = \boxed{84}$ .

Let,

√x be a

√y be c

√z be b

Therefore,

c²-ab=6 (eq 1)

b²-ac=-30 (eq 2)

a²-bc=42 (eq 3)

Subtracting eq 2 from eq 1

c²-b²+ac-ab=36

by factorising

(c-b)(a+b+c)=36

Subtracting eq 2 from eq 3

(a-b)(a+b+c)=72

Therefore, 2(c-b)=a-b

→c=(a+b)/2

Substituting c value in eq 1

((a+b)/2)²-ab=6

Solving for a and b we get

(a-b)²=24

→a-b=√24 as a>b

→a=√24+b (eq 4)

Substituting value of c in eq 3 give

(a-b)(2a+b)=84

Substituting 'a' from eq 4

√24(2√24+3b)=84

Solve b by substituting b=√z

z=6 and b=√6

By substituting the value of b we get x=54 and y=24

Therefore x+y+z =6+54+24=84.