Question

Solve each system in Exercises 5–18. ﻿{3(2x+y)+5z=−12(x−3y+4z)=−94(1+x)=−3(z−3y)\begin{cases}
3(2x + y) + 5z = -1 \
2(x - 3y + 4z) = -9 \
4(1 + x) = -3(z - 3y)
\end{cases}⎩⎨⎧​3(2x+y)+5z=−12(x−3y+4z)=−94(1+x)=−3(z−3y)​

Accepted Answer

First, write down the given system of equations clearly:

$$3(2x + y) + 5z = -1$$

$$2(x - 3y + 4z) = -9$$

$$4(1 + x) = -3(z - 3y)$$ Next, simplify each equation by distributing the constants inside the parentheses:

For the first equation: $$3 	imes 2x + 3 	imes y + 5z = -1$$ which simplifies to $$6x + 3y + 5z = -1.

$$For the second equation: $$2 	imes x - 2 	imes 3y + 2 	imes 4z = -9$$ which simplifies to $$2x - 6y + 8z = -9.

$$For the third equation: $$4 	imes 1 + 4 	imes x = -3 	imes z + 3 	imes 3y$$ which simplifies to $$4 + 4x = -3z + 9y. $$Rearrange the third equation to bring all terms to one side, making it consistent with the others:

$$4x - 9y + 3z = -4$$ Now, write the simplified system of linear equations:

$$6x + 3y + 5z = -1$$

$$2x - 6y + 8z = -9$$

$$4x - 9y + 3z = -4 To $$solve the system, choose a method such as substitution, elimination, or matrix methods (like Gaussian elimination). For example, you can:

- Use elimination to eliminate one variable between two equations,
- Then solve the resulting two-variable system,
- Substitute back to find the third variable.

Solve each system in Exercises 5–18. $\begin{cases}\)3(2x + y) + 5z = -1 \\2(x - 3y + 4z) = -9 \\4(1 + x) = -3(z - 3y)\(\end{cases}$

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Key Concepts

Systems of Linear Equations

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Methods for Solving Systems of Equations

Solve each system in Exercises 5–18. {3(2x+y)+5z=−12(x−3y+4z)=−94(1+x)=−3(z−3y)\(\begin{cases}\)3(2x + y) + 5z = -1 \\2(x - 3y + 4z) = -9 \\4(1 + x) = -3(z - 3y)\(\end{cases}\)⎩⎨⎧3(2x+y)+5z=−12(x−3y+4z)=−94(1+x)=−3(z−3y)