To find : the solution to the given system of linear equations
Answer to Problem 27E
The given system of equation has NO SOLUTION
Explanation of Solution
Given information : The system of equation is
Concept Involved:
Solution of a system of equation is the point which makes both the equation TRUE.
Graphically the solution to the system of equation is the point where the two lines meet.
Method of Elimination:
To use the method of elimination to solve a system of two linear equations in x and y, perform the following steps.
1. Obtain coefficients for x (or y) that differ only in sign by multiplying all
terms of one or both equations by suitably chosen constants.
2. Add the equations to eliminate one variable.
3. Solve the equation obtained in Step 2.
4. Back-substitute the value obtained in Step 3 into either of the original
equations and solve for the other variable.
5. Check that the solution satisfies each of the original equations.
The Number of Solutions of a Linear System
For a system of linear equations, exactly one of the following is true.
1. There is exactly one solution.
2. There are infinitely many solutions.
3. There is no solution.
Calculation:
Description | Steps | |
Label the given equations | ▶ 1st equation | |
▶ 2nd equation | ||
▶ 3rd equation | ||
In order to eliminate z multiply 2 from the 1st equation and add the result to the 2nd equation | ||
Label the new equation as 4th equation | ▶ 4th equation | |
Solving 4th equation By dividing 5 on both sides Simplify fraction on both sides | ||
Substitute 1 for x in 1st and 3rd equation, then add the 1st and 2nd equation |
Conclusion:
Since we end up with a false equation, there is NO SOLUTION to the system of equation
Chapter 7 Solutions
EBK PRECALCULUS W/LIMITS
- Find the length of the following curve. 3 1 2 N x= 3 -y from y 6 to y=9arrow_forward3 4/3 3213 + 8 for 1 ≤x≤8. Find the length of the curve y=xarrow_forwardGiven that the outward flux of a vector field through the sphere of radius r centered at the origin is 5(1 cos(2r)) sin(r), and D is the value of the divergence of the vector field at the origin, the value of sin (2D) is -0.998 0.616 0.963 0.486 0.835 -0.070 -0.668 -0.129arrow_forward
- 10 The hypotenuse of a right triangle has one end at the origin and one end on the curve y = Express the area of the triangle as a function of x. A(x) =arrow_forwardIn Problems 17-26, solve the initial value problem. 17. dy = (1+ y²) tan x, y(0) = √√3arrow_forwardcould you explain this as well as disproving each wrong optionarrow_forward
- could you please show the computation of this by wiresarrow_forward4 Consider f(x) periodic function with period 2, coinciding with (x) = -x on the interval [,0) and being the null function on the interval [0,7). The Fourier series of f: (A) does not converge in quadratic norm to f(x) on [−π,π] (B) is pointwise convergent to f(x) for every x = R П (C) is in the form - 4 ∞ +Σ ak cos(kx) + bk sin(kx), ak ‡0, bk ‡0 k=1 (D) is in the form ak cos(kx) + bk sin(kx), ak 0, bk 0 k=1arrow_forwardSolve the equation.arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning