To calculate:
The solution of the following linear equations by using any correct method:
Answer to Problem 24E
The solution of the
Explanation of Solution
Given information:
Calculation:
The given equations are as:
From above listed two equations, the coefficient of variable
Take addition of equation (1) and (2) to eliminate variable
Now calculate the value of variable
Hence, by solving two equations by elimination method get variable values as
Chapter 5 Solutions
BIG IDEAS MATH Integrated Math 1: Student Edition 2016
- By considering appropriate series expansions, e². e²²/2. e²³/3. .... = = 1 + x + x² + · ... when |x| < 1. By expanding each individual exponential term on the left-hand side the coefficient of x- 19 has the form and multiplying out, 1/19!1/19+r/s, where 19 does not divide s. Deduce that 18! 1 (mod 19).arrow_forwardProof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.arrow_forwardBy considering appropriate series expansions, ex · ex²/2 . ¸²³/³ . . .. = = 1 + x + x² +…… when |x| < 1. By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x 19 has the form 1/19!+1/19+r/s, where 19 does not divide s.arrow_forward
- Let 1 1 r 1+ + + 2 3 + = 823 823s Without calculating the left-hand side, prove that r = s (mod 823³).arrow_forwardFor each real-valued nonprincipal character X mod 16, verify that L(1,x) 0.arrow_forward*Construct a table of values for all the nonprincipal Dirichlet characters mod 16. Verify from your table that Σ x(3)=0 and Χ mod 16 Σ χ(11) = 0. x mod 16arrow_forward
- For each real-valued nonprincipal character x mod 16, verify that A(225) > 1. (Recall that A(n) = Σx(d).) d\narrow_forward24. Prove the following multiplicative property of the gcd: a k b h (ah, bk) = (a, b)(h, k)| \(a, b)' (h, k) \(a, b)' (h, k) In particular this shows that (ah, bk) = (a, k)(b, h) whenever (a, b) = (h, k) = 1.arrow_forward20. Let d = (826, 1890). Use the Euclidean algorithm to compute d, then express d as a linear combination of 826 and 1890.arrow_forward
- Let 1 1+ + + + 2 3 1 r 823 823s Without calculating the left-hand side, Find one solution of the polynomial congruence 3x²+2x+100 = 0 (mod 343). Ts (mod 8233).arrow_forwardBy considering appropriate series expansions, prove that ez · e²²/2 . e²³/3 . ... = 1 + x + x² + · ·. when <1.arrow_forwardProve that Σ prime p≤x p=3 (mod 10) 1 Р = for some constant A. log log x + A+O 1 log x ,arrow_forward
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