Question

fluency and skills practic
lesson 13
solving systems of li
elimination
find the solution to each system of ec
1 ( 4x - 12y = -8 )
( -3x + 12y = 12 )
3 ( 6x + 3y = 3 )
( 3x - y = 4 )

Explanation:

Problem 1:

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

We have the system:
\[

$$\begin{cases} 4x - 12y = -8 \\ -3x + 12y = 12 \end{cases}$$

\]
Adding the left - hand sides and the right - hand sides of the two equations:
\((4x-12y)+(-3x + 12y)=-8 + 12\)
Simplify the left - hand side: \(4x-12y-3x + 12y=x\), and the right - hand side: \(4\). So we get \(x = 4\).

Step2: Substitute \( x = 4 \) into one of the original equations to find \( y \)

Let's substitute \( x = 4 \) into the first equation \(4x-12y=-8\).
Substitute \( x = 4 \) into \(4x-12y=-8\):
\(4\times4-12y=-8\)
\(16-12y=-8\)
Subtract 16 from both sides: \(-12y=-8 - 16=-24\)
Divide both sides by \(-12\): \(y=\frac{-24}{-12}=2\)

Problem 3:

Step1: Make the coefficients of \( y \) opposite (or equal)

We have the system:
\[

$$\begin{cases} 6x+3y = 3 \\ 3x-y = 4 \end{cases}$$

\]
Multiply the second equation \(3x - y=4\) by 3, we get \(9x-3y = 12\)

Step2: Add the two equations to eliminate \( y \)

Now our system becomes:
\[

$$\begin{cases} 6x+3y = 3 \\ 9x-3y = 12 \end{cases}$$

\]
Add the two equations: \((6x + 3y)+(9x-3y)=3 + 12\)
Simplify the left - hand side: \(6x+3y+9x-3y = 15x\), and the right - hand side: \(15\)
So \(15x=15\), divide both sides by 15, we get \(x = 1\)

Step3: Substitute \( x = 1 \) into one of the original equations to find \( y \)

Substitute \( x = 1 \) into the second equation \(3x-y = 4\)
\(3\times1-y = 4\)
\(3-y = 4\)
Subtract 3 from both sides: \(-y=4 - 3 = 1\)
Multiply both sides by \(-1\): \(y=-1\)

Answer:

s:

1. The solution of the system \(

$$\begin{cases}4x - 12y=-8\\-3x + 12y = 12\end{cases}$$

\) is \(x = 4,y = 2\)

1. The solution of the system \(

$$\begin{cases}6x+3y = 3\\3x-y = 4\end{cases}$$

\) is \(x = 1,y=-1\)