Question

section 5.2 - systems of linear equations in three variables and applications question 4 of 11 (1 point) | question attempt: 1 of unlimited solve the system of equations. if the system does not have one unique solution, determine the number of solutions to the system. $4x - 3y = 28$ $7x + 3z = 19$ $7y + 4z = -40$ - the system has one solution. the solution set is ${(square, square, square)}$. - the system has no solution. - the system is inconsistent. - the equations are dependent. - the system has infinitely many solutions. - the system is inconsistent. - the equations are dependent.

Explanation:

Step1: Express \(x\) from the second equation

From \(7x + 3z=19\), we get \(x=\frac{19 - 3z}{7}\).

Step2: Express \(y\) from the third equation

From \(7y+4z=-40\), we get \(y=\frac{-40 - 4z}{7}\).

Step3: Substitute \(x\) and \(y\) into the first equation

Substitute \(x=\frac{19 - 3z}{7}\) and \(y=\frac{-40 - 4z}{7}\) into \(4x-3y = 28\).

$$
LATEXBLOCK0
$$

Now, substitute \(z = - 5\) into \(x=\frac{19 - 3z}{7}\), \(x=\frac{19-3\times(-5)}{7}=\frac{19 + 15}{7}=4\).
Substitute \(z=-5\) into \(y=\frac{-40 - 4z}{7}\), \(y=\frac{-40-4\times(-5)}{7}=\frac{-40 + 20}{7}=-4\).

Answer:

The system has one solution. The solution set is \(\{(4,-4,-5)\}\).