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A box containing a total of 179 copies of two different paperback books was shipped to marci’s school. The total weight of the books was 128 pounds. If the weight of each of the first paperbacks was two-thirds of a pound and the weight of each of the second paperbacks was three-fourths of a pound, which statements are true? check all that apply. The system of equations is x + y = 179 and two-thirds x + three-fourths y = 128. The system of equations is x + y = 128 and two-thirds x + three-fourths y = 179. To eliminate the x-variable from the equations, you can multiply the equation with the fractions by 3 and leave the other equation as it is. To eliminate the y-variable from the equations, you can multiply the equation with the fractions by –4 and multiply the other equation by 3. There are 104 copies of one book and 24 copies of the other. There are 93 copies of one book and 35 copies of the other. There are 104 copies of one book and 75 copies of the other. There are 93 copies of one book and 86 copies of the other.

Sagot :

Adioabiolaid

The system of equation to represent the situation is

x + y = 179

2/3x + 3/4y = 128

How to solve simultaneous equation

  • Total copies = 179
  • Total weight = 128 pounds

let,

weight of first paperbacks, x = 2/3 pounds

weight of second paperbacks, y = 3/4 pounds

x + y = 179 (1)

2/3x + 3/4y = 128 (2)

  • multiply (1) by 3 and (2) by -4

3x + 3y = 537

-4(2/3x) + -4(3/4y) = -512

3x + 3y = 537

-8/3x - 3y = -512

  • Add both equations to eliminate y

3x + (-8/3x) = 537 + (-512)

3x - 8/3x = 537 - 512

(9-8)/3x = 25

1/3x = 25

x = 25 ÷ 1/3

x = 25 × 3

x = 75

  • Substitute x = 25 into (1)

x + y = 179

75 + y = 179

y = 179 - 75

y = 104

  • The system of equations is x + y = 179 and two-thirds x + three-fourths y = 128.

  • To eliminate the y-variable from the equations, you can multiply the equation with the fractions by –4 and multiply the other equation by 3.

  • There are 104 copies of one book and 75 copies of the other.

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