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Solve the following system of linear equations: 3x + 4y = 10 and 2x + 5y = 9.

Mathematics

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Answer

x=2,y=1x=2, y=1

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The system of equations for is:

3x+4y=10(1)3x + 4y = 10 \quad (1) 2x+5y=9(2)2x + 5y = 9 \quad (2)

  1. Multiply the equations to make the coefficients of xx equal.

    Multiply equation (1) by 2 and equation (2) by 3:

    2×(3x+4y)=2×10  ⟹  6x+8y=20(3)2 \times (3x + 4y) = 2 \times 10 \implies 6x + 8y = 20 \quad (3) 3×(2x+5y)=3×9  ⟹  6x+15y=27(4)3 \times (2x + 5y) = 3 \times 9 \implies 6x + 15y = 27 \quad (4)

  2. Subtract equation (3) from equation (4) to eliminate xx.

    (6x+15y)−(6x+8y)=27−20(6x + 15y) - (6x + 8y) = 27 - 20 6x−6x+15y−8y=76x - 6x + 15y - 8y = 7 7y=77y = 7

  3. Solve for yy.

    7y7=77\frac{7y}{7} = \frac{7}{7} y=1y = 1

  4. Substitute the value of yy into equation (1) to find xx.

    3x+4(1)=103x + 4(1) = 10 3x+4=103x + 4 = 10 3x=10−43x = 10 - 4 3x=63x = 6

  5. Solve for xx.

    3x3=63\frac{3x}{3} = \frac{6}{3} x=2x = 2

    The solution to the system of equations is x=2x=2 and y=1y=1.

    x=2,y=1x=2, y=1
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