Two Victorious Ways to Solve a Challenging System of Equations

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In this algebraic video, we'll explore two powerful methods to tackle a challenging system of equations. Whether you're a math enthusiast or a student seeking to improve your problem-solving skills, these strategies will enhance your understanding and boost your confidence. We'll break down each method step-by-step, providing clear explanations and examples to ensure you grasp the concepts. Don't miss out on mastering these valuable techniques!
Topics covered:
System of equations
Algebra
Problem solving
Algebraic identities
Algebraic manipulations
Solving systems of equations
Factorization
Math enthusiast
Math tutorial
Math Olympiad
Math Olympiad Preparation
Time-stamps:
0:00 Introduction
0:40 Method-1
2:17 Algebraic identities
7:10 Quadratic equations
10:44 Method-2
11:50 Solving system of equations
12:01 Pascal's triangle
14:31 Binomial expansion
19:20 Discriminant
19:45 Solutions
#mathtutorial #systemofequations #problemsolving #mathhelp #algebra #learnmath #mathtricks #studytips #education #solvingequations #mathskills #mathenthusiast #stem #math
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Thanks for watching!
@infyGyan

[@tejpalsingh366]

(x;y)=(4; 1); (1; 4)

[@mohammedsaysrashid3587]

A wonderful explanation and nice solution....thanks for sharing...finally ( x,y) ( 4,1) ( 1,4)

[@thichhochoi766]

Once you got sqrt(x)+sqrt(y) = 3 and sqrt(xy) = (2,7), you can use sum and product rule to solve for x and y.

[@alibhukoo5400]

Either X=4 y=1 or x=1 y=4

[@abcekkdo3749]

(X=4,y=1)
(X=1,y=4)

[@jim2376]

By inspection (4,1) and (1,4). 33 and 3 tell us we're dealing the relatively small numbers.

[@user-ny6jf9is3t]

(x,y)=(1,4) ή (4,1) απλη σκεψη αλλα πολλες πραξεις.

[@woobjun2582]

By letting
sqrt(x) =a and sqrt(y) =b,
the given becomes
a^5 +b^5 =33 ..(1)
a +b = 3 ..(2)
By cubing (2),
a^3 +b^3 +3ab(a+b) =27;
a^3 +b^3 = 27 - 9ab ..(3).
By powering 5 for (2),
a^5 +b^5
+5ab(a^3 +b^3)
+10[(ab)^2](a +b) =243.
Substituting (1) thru (3)
33 +5ab(27 -9ab)
+10(3)(ab)^2 = 243;
33 +135ab -45(ab)^2
+30(ab)^2 =243,
that is
15(ab)^2 -135ab +210 =0;
(ab)^2 -9ab +14 =0;
(ab)^2 -7ab -2ab +14 =0;
ab(ab -7) -2(ab -7) =0;
(ab -7)(ab -2) =0,
and thus
ab =7 or ab =2
which leads to solve
2 systems of eqns
a +b =3 and ab =7 ..(S1)
a +b =3 and ab =2 ..(S2)
Solving (S1)
a^2 -3a +7 =0
results in cmplx sols.
Then to solve (S2)
a^2 -3a +2 =0;
(a -1)(a -2) =0,
thus
a = 1, 2 and b = 2, 1
Recall a^2 =x & b^2 =y,
(x, y) = (1, 4), (4, 1).

[@ashokdubey8415]

solved in similar manner

[@woobjun2582]

@@ashokdubey8415 Let's go to find another way as the lecturer mentioned 2 ways^.^

[@RB-4]

X=1
Y=4

[@walterwen2975]

Solve a Challenging System of Equations:
x²(√x) + y²(√y) = 33, √x + √y = 3; x, y = ?
3 > √x > √y > 0 or 3 > √y > √x > 0
Let: u = √x, v = √y; u + v = 3, x²(√x) + y²(√y) = u⁵ + v⁵ = 33
u³ + v³ = (u + v)³ - 3uv(u + v) = 3³ - 3uv(3) = 9(3 - uv)
u² + v² = (u + v)² - 2uv = 3² - 2uv = 9 - 2uv
(u² + v²)(u³ + v³) = u⁵ + v⁵ +(u + v)(uv)² = 33 + 3(uv)² = 9(3 - uv)(9 - 2uv)
11 + (uv)² = 3[27 - 15uv + 2(uv)²] = 81 - 45uv + 6(uv)²
5(uv)² - 45uv + 70 = 0, (uv)² - 9uv + 14 = (uv - 2)(uv - 7) = 0
uv - 2 = 0; uv = 2 or uv - 7 = 0; uv = 7
uv = 2: (u - v)² = (u + v)² - 4uv = 3² - 4(2) = 1; u - v = ± 1, u + v = 3
2u = 4; u = 2, v = 1, or 2u = 2; u = 1, v = 2
u = √x = 2, x = 4, v = √y = 1, y = 1 or x = 1, y = 4
uv = 7: (u - v)² = (u + v)² - 4uv = 3² - 4(7) = - 17 < 0; Rejected
Answer check:
x = 4, y = 1 or x = 1, y = 4
x²(√x) + y²(√y) = 4²(√4) + 1 = 1 + 4²(√4) = 1 + 32 = 33; Confirmed
√x + √y = √4 + 1 = 1 + √4 = 2 + 1= 3; Confirmed
Final answer:
x = 4, y = 1 or x = 1, y = 4

[@infyGyan]

Spectacular!