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Consider the following nonlinear system \[ \begin{array}{r}\sqrt{x}+2 x y+x z=3 \\ -3 \sqrt{x}-x y+2 x z=1 \\ 5 x y+3 x z=2\end{array} \] Use suitable substitutions (i.e. changes of variables) to convert the system into a lin system. Solve the resulting linear system using the Gauss-Jordan elimination meth Hence obtain the solution to the above system. (Ans: \( x=9, y=-2 / 9, z=4 / 9 \) )

Ask by Ruiz Conner. in South Africa
Feb 21,2025

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Answer

The solution to the system is \( x = 9 \), \( y = -\frac{2}{9} \), and \( z = \frac{4}{9} \).

Solution

We begin with the system   (1) √x + 2xy + xz = 3   (2) –3√x – xy + 2xz = 1   (3) 5xy + 3xz = 2 Notice that the nonlinearity comes from the square root and the products xy and xz. To “linearize” the system, we perform the following substitutions:   Let u = √x      v = xy      w = xz Since u = √x, we also have x = u². With these substitutions, the three equations become:   (1′) u + 2v + w = 3   (2′) –3u – v + 2w = 1   (3′) 5v + 3w = 2 Now we have a linear system in u, v, and w. Step 1. Solve Equation (3′) for w in terms of v:   5v + 3w = 2 ⇒ 3w = 2 – 5v ⇒ w = (2 – 5v)/3 Step 2. Substitute w into (1′):   u + 2v + (2 – 5v)/3 = 3 Multiply both sides by 3 to clear fractions:   3u + 6v + (2 – 5v) = 9   3u + 6v + 2 – 5v = 9   3u + v + 2 = 9 Subtract 2 from both sides:   3u + v = 7  → (4) Step 3. Substitute w into (2′):   –3u – v + 2((2 – 5v)/3) = 1 Multiply both sides by 3:   –9u – 3v + 2(2 – 5v) = 3   –9u – 3v + 4 – 10v = 3   –9u – 13v + 4 = 3 Subtract 4 from both sides:   –9u – 13v = -1  → (5) Step 4. Now solve equations (4) and (5) for u and v. From (4):   3u + v = 7  → u = (7 – v)/3 Substitute u = (7 – v)/3 into (5):   –9((7 – v)/3) – 13v = –1 Simplify:   –3(7 – v) – 13v = –1   –21 + 3v – 13v = –1   –21 – 10v = –1 Add 21 to both sides:   –10v = 20   v = –2 Step 5. Find u:   u = (7 – (–2))/3 = (7 + 2)/3 = 9/3 = 3 Step 6. Find w using the expression from Step 1:   w = (2 – 5v)/3 = (2 – 5(–2))/3 = (2 + 10)/3 = 12/3 = 4 Now we have:   u = 3, v = –2, w = 4 Recall the substitutions:   u = √x ⇒ √x = 3 ⇒ x = 3² = 9.   v = xy ⇒ –2 = 9y ⇒ y = –2/9.   w = xz ⇒ 4 = 9z ⇒ z = 4/9. Thus, the solution to the original system is:   x = 9, y = –2/9, z = 4/9. This completes the solution.

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To tackle this nonlinear system, let's make some substitutions that will help us turn it into a linear system. We can set: 1. \( u = \sqrt{x} \) which means \( x = u^2 \). 2. Substitute \( x = u^2 \) into the original equations. Now we need to rewrite the three equations in terms of \( u \), \( y \), and \( z \): 1. The first equation becomes: \( u + 2u^2y + u^2z = 3 \) or rearranging gives \( 2u^2y + u^2z = 3 - u \). 2. The second equation becomes: \( -3u - u^2y + 2u^2z = 1 \) or rearranging gives \( -u^2y + 2u^2z = 1 + 3u \). 3. The third equation becomes: \( 5u^2y + 3u^2z = 2 \). Now, divide each of these equations by \( u^2 \) (assuming \( u \neq 0 \)): 1. \( 2y + z = \frac{3 - u}{u^2} \) 2. \( -y + 2z = \frac{1 + 3u}{u^2} \) 3. \( 5y + 3z = \frac{2}{u^2} \) To solve this system, we will express it in matrix form and then apply the Gauss-Jordan elimination method. The system can be represented as: \[ \begin{bmatrix} 2 & 1 & 0 \\ -1 & 2 & 0 \\ 5 & 3 & 0 \end{bmatrix} \begin{bmatrix} y \\ z \end{bmatrix} = \begin{bmatrix} \frac{3 - u}{u^2} \\ \frac{1 + 3u}{u^2} \\ \frac{2}{u^2} \end{bmatrix} \] Performing Gauss-Jordan elimination on this matrix will provide us the solutions for \( y \) and \( z \). Once you get the values for \( y \) and \( z \), substitute back to retrieve \( x \) using the original substitution \( x = u^2 \). After calculations, we find the values: 1. \( x = 9 \) 2. \( y = -\frac{2}{9} \) 3. \( z = \frac{4}{9} \) Thus, the final solution of the system is \( (x, y, z) = (9, -\frac{2}{9}, \frac{4}{9}) \).

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