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Determine if the following pairs of linear equations are consistent or inconsistent.

Mathematics

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Step-by-step solution

12 steps

To determine if a pair of linear equations is consistent or inconsistent, we compare the ratios of their coefficients. For a system of equations:
a1x+b1y+c1=0a_1x + b_1y + c_1 = 0
a2x+b2y+c2=0a_2x + b_2y + c_2 = 0

  • If a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}, the system is consistent (unique solution).
  • If a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}, the system is consistent (infinitely many solutions).
  • If a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}, the system is inconsistent (no solution).

Let's solve the visible parts of the problem.

Part i· 4 steps

The given equations are:
3x+2y=5  ⟹  3x+2y−5=03x + 2y = 5 \implies 3x + 2y - 5 = 0
2x−3y=7  ⟹  2x−3y−7=02x - 3y = 7 \implies 2x - 3y - 7 = 0

  1. Identify the coefficients.

    Here, a1=3a_1 = 3, b1=2b_1 = 2, c1=−5c_1 = -5
    And a2=2a_2 = 2, b2=−3b_2 = -3, c2=−7c_2 = -7

  2. Calculate the ratios.

    a1a2=32\frac{a_1}{a_2} = \frac{3}{2} b1b2=2−3=−23\frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3} c1c2=−5−7=57\frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7}
  3. Compare the ratios.

    Since 32≠−23\frac{3}{2} \neq -\frac{2}{3}, we have a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}.

  4. Conclude consistency.

    Therefore, the pair of equations is consistent.

Part iii· 4 steps

The given equations are:
32x+53y=7  ⟹  32x+53y−7=0\frac{3}{2}x + \frac{5}{3}y = 7 \implies \frac{3}{2}x + \frac{5}{3}y - 7 = 0
9x−10y=14  ⟹  9x−10y−14=09x - 10y = 14 \implies 9x - 10y - 14 = 0

  1. Identify the coefficients.

    Here, a1=32a_1 = \frac{3}{2}, b1=53b_1 = \frac{5}{3}, c1=−7c_1 = -7
    And a2=9a_2 = 9, b2=−10b_2 = -10, c2=−14c_2 = -14

  2. Calculate the ratios.

    a1a2=329=32×9=318=16\frac{a_1}{a_2} = \frac{\frac{3}{2}}{9} = \frac{3}{2 \times 9} = \frac{3}{18} = \frac{1}{6} b1b2=53−10=53×(−10)=5−30=−16\frac{b_1}{b_2} = \frac{\frac{5}{3}}{-10} = \frac{5}{3 \times (-10)} = \frac{5}{-30} = -\frac{1}{6} c1c2=−7−14=12\frac{c_1}{c_2} = \frac{-7}{-14} = \frac{1}{2}
  3. Compare the ratios.

    Since 16≠−16\frac{1}{6} \neq -\frac{1}{6}, we have a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}.

  4. Conclude consistency.

    Therefore, the pair of equations is consistent.

Part v· 4 steps

The given equations are:
−x+2y=8  ⟹  −x+2y−8=0-x + 2y = 8 \implies -x + 2y - 8 = 0
2x+3y=12  ⟹  2x+3y−12=02x + 3y = 12 \implies 2x + 3y - 12 = 0

  1. Identify the coefficients.

    Here, a1=−1a_1 = -1, b1=2b_1 = 2, c1=−8c_1 = -8
    And a2=2a_2 = 2, b2=3b_2 = 3, c2=−12c_2 = -12

  2. Calculate the ratios.

    a1a2=−12\frac{a_1}{a_2} = \frac{-1}{2} b1b2=23\frac{b_1}{b_2} = \frac{2}{3} c1c2=−8−12=23\frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}
  3. Compare the ratios.

    Since −12≠23-\frac{1}{2} \neq \frac{2}{3}, we have a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}.

  4. Conclude consistency.

    Therefore, the pair of equations is consistent.

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