Question

1. find the solution of the system of equations.

$-3x - 2y = 36$
$3x - 5y = 6$

1. find the solution of the system of equations.

$10x + y = -45$
$2x + y = -5$

1. solve the system by substitution.

$-5x + 3y = 2$
$y = 2x$

1. solve the system by substitution.

$-x + 3y = 15$
$y = 2x$

1. solve the system by substitution.

$-5y = x$
$3x + 9y = 24$

Explanation:

Problem 1

Step1: Add the two equations to eliminate \(x\)

We have the system:

$$
LATEXBLOCK0
$$

Adding the two equations: \((-3x - 2y)+(3x - 5y)=36 + 6\)
Simplify: \(-7y=42\)

Step2: Solve for \(y\)

Divide both sides by \(-7\): \(y=\frac{42}{-7}=-6\)

Step3: Substitute \(y = - 6\) into one of the original equations to find \(x\)

Let's use the second equation \(3x-5y = 6\). Substitute \(y=-6\):
\(3x-5\times(-6)=6\)
\(3x + 30=6\)
Subtract 30 from both sides: \(3x=6 - 30=-24\)
Divide by 3: \(x=\frac{-24}{3}=-8\)

Step1: Subtract the second equation from the first to eliminate \(y\)

We have the system:

$$
LATEXBLOCK0
$$

Subtract the second equation from the first: \((10x + y)-(2x + y)=-45-(-5)\)
Simplify: \(8x=-40\)

Step2: Solve for \(x\)

Divide both sides by 8: \(x=\frac{-40}{8}=-5\)

Step3: Substitute \(x = - 5\) into one of the original equations to find \(y\)

Let's use the second equation \(2x + y=-5\). Substitute \(x=-5\):
\(2\times(-5)+y=-5\)
\(-10 + y=-5\)
Add 10 to both sides: \(y=-5 + 10 = 5\)

Step1: Substitute \(y = 2x\) into the first equation

We have the system:

$$
LATEXBLOCK0
$$

Substitute \(y = 2x\) into \(-5x + 3y=2\):
\(-5x+3\times(2x)=2\)

Step2: Simplify and solve for \(x\)

Simplify: \(-5x+6x=2\)
\(x = 2\)

Step3: Substitute \(x = 2\) into \(y = 2x\) to find \(y\)

\(y=2\times2 = 4\)

Answer:

The solution is \(x=-8,y = - 6\)

Problem 2