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Solve the following simultaneous equations: 3/4 p + 2q = 3; p + 2q = 6. (c) Adama and Janita shared X cups of rice in the ratio 2:3. If Janita's share is 270 cups, find the value of X.

Mathematics

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2 parts
b
12
-3
c

450 cups450 \text{ cups}

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here are the solutions to the questions:

Part b· 4 steps

Solve the following simultaneous equations

34p+2q=3(1)\frac{3}{4}p + 2q = 3 \quad \text{(1)} p+2q=6(2)p + 2q = 6 \quad \text{(2)}

  1. From equation (2), express 2q2q in terms of pp.

    2q=6−p(3)2q = 6 - p \quad \text{(3)}

  2. Substitute equation (3) into equation (1).

    34p+(6−p)=3\frac{3}{4}p + (6 - p) = 3

  3. Solve for pp.

    34p+6−p=3\frac{3}{4}p + 6 - p = 3 6−3=p−34p6 - 3 = p - \frac{3}{4}p 3=44p−34p3 = \frac{4}{4}p - \frac{3}{4}p 3=14p3 = \frac{1}{4}p p=3×4p = 3 \times 4 p=12p = 12

  4. Substitute the value of pp back into equation (3) to find qq.

    2q=6−122q = 6 - 12 2q=−62q = -6 q=−62q = \frac{-6}{2} q=−3q = -3

    The solution is p=12p = \boxed{12} and q=−3q = \boxed{-3}.

Part c

450 cups450 \text{ cups}

Adama and Janita shared X cups of rice in the ratio 2:3. If Janita's share is 270 cups, find the value of X.

  1. Determine the total number of ratio parts.

    The ratio for Adama : Janita is 2 : 3.
    Total parts = 2+3=52 + 3 = 5 parts.

  2. Determine the fraction of the total share that Janita received.

    Janita's share corresponds to 3 parts out of 5 total parts, so Janita received 35\frac{3}{5} of the total rice.

  3. Set up an equation using Janita's share and solve for X.

    35×X=270\frac{3}{5} \times X = 270 X=270×53X = 270 \times \frac{5}{3} X=90×5X = 90 \times 5 X=450X = 450

    The value of X is 450 cups\boxed{450 \text{ cups}}.

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