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) $$ y-2=2(x-1)^{2} $$

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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.

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(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).

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(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).

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(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).

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(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).

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(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.

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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.
(a) $$ x = 0, y = 4 $$ or $$ x = 2, y = -8 $$ (b) $$ x = 3, y = -2 $$ or $$ x = -5, y = -18 $$ (c) $$ x = 0, y = 6 $$ or $$ x = 6, y = 0 $$ (d) $$ x = 4, y = 2 $$ or $$ x = -20, y = -14 $$ (e) $$ y = 2(x - 1)^2 + 2 $$

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) \( y-2=2(x-1)^{2} \)

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