1) Sutve fir \( x \) and \( y \) in each of the following e (a) \( y+6 x=4 \) and \( 2 y=5 x^{2}-22 x+8 \) b) \( 2 x-y=8 \) and \( y=x^{2}+4 x-23 \) (d) \( y+x=6 \) and \( x^{2}+y^{2}=36 \) d) \( 2 x-3 y=2 \) and \( x^{2}-2 y^{2}=8 \) e) \( y-2=2(x-1)^{2} \)

We wish to “solve for x and y” in each case. In some cases the answer is a pair (or pairs) of numbers; in one case (part (e)) there is only one equation so the answer is an expression of y in terms of x. One acceptable solution is given below. ────────────────────────────── (a) Solve   (1) y + 6x = 4   (2) 2y = 5x² – 22x + 8 Step 1. Solve (1) for y:   y = 4 – 6x. Step 2. Substitute into (2):   2(4 – 6x) = 5x² – 22x + 8.   8 – 12x = 5x² – 22x + 8. Step 3. Subtract 8 on both sides:   8 – 12x – 8 = 5x² – 22x  ⟹  –12x = 5x² – 22x. Step 4. Bring all terms to one side:   0 = 5x² – 22x + 12x = 5x² – 10x. Step 5. Factor:   5x² – 10x = 5x(x – 2) = 0. Thus, either x = 0 or x = 2. Step 6. Find y from y = 4 – 6x:   If x = 0, then y = 4.   If x = 2, then y = 4 – 12 = –8. Answer (a): (x, y) = (0, 4) or (2, –8). ────────────────────────────── (b) Solve   (1) 2x – y = 8   (2) y = x² + 4x – 23 Step 1. From (1) express y:   y = 2x – 8. Step 2. Set equal to (2):   2x – 8 = x² + 4x – 23. Step 3. Bring all terms to one side:   0 = x² + 4x – 23 – 2x + 8 = x² + 2x – 15. Step 4. Solve the quadratic x² + 2x – 15 = 0.   The discriminant is D = 2² – 4(1)(–15) = 4 + 60 = 64.   Thus, x = [–2 ± √64]⁄2 = (–2 ± 8)/2.   So, x = 3 or x = –5. Step 5. Now find y using y = 2x – 8:   If x = 3, then y = 6 – 8 = –2.   If x = –5, then y = –10 – 8 = –18. Answer (b): (x, y) = (3, –2) or (–5, –18). ────────────────────────────── (c) Solve   (1) x + y = 6   (2) x² + y² = 36 Step 1. Express y from (1):   y = 6 – x. Step 2. Substitute into (2):   x² + (6 – x)² = 36.   x² + (36 – 12x + x²) = 36.   2x² – 12x + 36 = 36. Step 3. Subtract 36:   2x² – 12x = 0 ⟹ 2x(x – 6) = 0. Thus either x = 0 or x = 6. Step 4. Find y:   If x = 0, then y = 6.   If x = 6, then y = 0. Answer (c): (x, y) = (0, 6) or (6, 0). ────────────────────────────── (d) Solve   (1) 2x – 3y = 2   (2) x² – 2y² = 8 Step 1. Solve (1) for x:   2x = 2 + 3y ⟹ x = (2 + 3y)/2. Step 2. Substitute into (2):   [(2 + 3y)/2]² – 2y² = 8.   (4 + 12y + 9y²)/4 – 2y² = 8. Step 3. Multiply both sides by 4:   4 + 12y + 9y² – 8y² = 32 ⟹ 4 + 12y + y² = 32. Step 4. Rearranging:   y² + 12y + 4 – 32 = 0 ⟹ y² + 12y – 28 = 0. Step 5. Solve the quadratic:   The discriminant is D = 144 + 112 = 256.   Thus, y = [–12 ± √256]⁄2 = (–12 ± 16)/2.   So, y = 2 or y = –14. Step 6. Find corresponding x:   For y = 2: x = (2 + 3·2)/2 = (2 + 6)/2 = 4.   For y = –14: x = (2 + 3·(–14))/2 = (2 – 42)/2 = –40/2 = –20. Answer (d): (x, y) = (4, 2) or (–20, –14). ────────────────────────────── (e) Solve   y – 2 = 2(x – 1)² This is a single equation (not a system); it represents the equation of a parabola. One common way to “solve” it is to express y in terms of x. Simply add 2 to both sides:   y = 2(x – 1)² + 2. Thus the solution set is all pairs (x, y) with x arbitrary and y given by y = 2(x – 1)² + 2. ────────────────────────────── Summary of Answers (a) (x, y) = (0, 4) or (2, –8). (b) (x, y) = (3, –2) or (–5, –18). (c) (x, y) = (0, 6) or (6, 0).  [This was labeled (d) in the original text.] (d) (x, y) = (4, 2) or (–20, –14).  [Also labeled (d) originally.] (e) The equation is equivalent to y = 2(x – 1)² + 2 (with x any real number). Any answers equivalent to the above are correct.