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For the following system of equations in echelon form, tell how many solutions there are in nonnegative integers: 2x + 3y + 3z = 80, y - 3z = 10?

User DMabulage
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Final answer:

The system of equations in echelon form, 2x + 3y + 3z = 80 and y - 3z = 10, has a single solution in nonnegative integers: x = 25, y = 10, z = 0.

Step-by-step explanation:

This system of equations is in echelon form, which means it is already simplified and organized.

From the second equation, we have y - 3z = 10. We can isolate y by adding 3z to both sides, giving us y = 10 + 3z.

Substituting this value of y into the first equation, we get 2x + 3(10 + 3z) + 3z = 80. Simplifying this equation gives us 2x + 33z = 50. Rearranging this equation, we have 33z = 50 - 2x.

Since x and z must be nonnegative integers, we can analyze the possible values of z. From the available information, we know that z is approximately -0.33 and -1.67. However, since z must be a nonnegative integer, the only possible value for z is 0.

Substituting z = 0 into 33z = 50 - 2x, we have 33(0) = 50 - 2x, which simplifies to 0 = 50 - 2x. Solving for x, we get x = 25. Now that we have the values of x and z, we can substitute them into y = 10 + 3z to find y. Substituting z = 0, we have y = 10 + 3(0), which simplifies to y = 10.

Therefore, the system of equations has a single solution in nonnegative integers: x = 25, y = 10, z = 0.

User Marwan Salim
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