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Solve the following equations for x and y: 2.3.1 Solve for x: (3^x)(3^(x−2)) = 27 2.3.2 Solve for x: -2x2 + x = -3x(-2x+1) 2.4 Solve algebraically for x and y: 1) 2x + 3y + 4 = 0 2) x + 2y − 5 = 0

Mathematics

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Here are the solutions to the problems:

2.3.1. Solve for xx:

(3x)(3x−2)=27(3^x)(3^{x-2}) = 27
  1. Apply the exponent rule am⋅an=am+na^m \cdot a^n = a^{m+n} on the left side and express the right side as a power of 3.

    3x+(x−2)=333^{x + (x-2)} = 3^3 32x−2=333^{2x-2} = 3^3
  2. Since the bases are equal, equate the exponents.

    2x−2=32x - 2 = 3
  3. Solve for xx.

    2x=3+22x = 3 + 2 2x=52x = 5 x=52x = \frac{5}{2}

    The solution is x=52\boxed{x = \frac{5}{2}}.

    2.3.2. Solve for xx:

    −2x2+x=−3x(−2x+1)-2x^2 + x = -3x(-2x+1)
  4. Expand the right side of the equation.

    −2x2+x=(−3x)(−2x)+(−3x)(1)-2x^2 + x = (-3x)(-2x) + (-3x)(1) −2x2+x=6x2−3x-2x^2 + x = 6x^2 - 3x
  5. Move all terms to one side to form a standard quadratic equation.

    0=6x2+2x2−3x−x0 = 6x^2 + 2x^2 - 3x - x 0=8x2−4x0 = 8x^2 - 4x
  6. Factor out the common term.

    0=4x(2x−1)0 = 4x(2x - 1)
  7. Set each factor equal to zero and solve for xx.

    4x=0or2x−1=04x = 0 \quad \text{or} \quad 2x - 1 = 0 x=0or2x=1x = 0 \quad \text{or} \quad 2x = 1 x=0orx=12x = 0 \quad \text{or} \quad x = \frac{1}{2}

    The solutions are x=0 or x=12\boxed{x = 0 \text{ or } x = \frac{1}{2}}.

    2.4. Solve algebraically for xx and yy in the following equations:

    1. 2x+3y+4=02x + 3y + 4 = 0
    2. x+2y−5=0x + 2y - 5 = 0
  8. Isolate xx from equation (2).

    x=−2y+5(Equation 3)x = -2y + 5 \quad \text{(Equation 3)}
  9. Substitute Equation 3 into Equation 1.

    2(−2y+5)+3y+4=02(-2y + 5) + 3y + 4 = 0
  10. Simplify and solve for yy.

    −4y+10+3y+4=0-4y + 10 + 3y + 4 = 0 −y+14=0-y + 14 = 0 −y=−14-y = -14 y=14y = 14
  11. Substitute the value of yy back into Equation 3 to find xx.

    x=−2(14)+5x = -2(14) + 5 x=−28+5x = -28 + 5 x=−23x = -23

    The solution is x=−23,y=14\boxed{x = -23, y = 14}.

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