Question

directions: homework

1. **solve the system of equations by using a table.

$y = 2x - 1$
$y = -x + 5$
table a
work space:
table b:
work space:
solution:

1. **consider the system given by $\

$$\begin{cases}y = 8x + 2\\\\y = 7x + 3\\end{cases}$$

$
is $(1, 10)$ a solution to the system?
justify your answer.

1. **consider the system given by

$\

$$\begin{cases}3x - 2y = -17\\\\5x + y = 13\\end{cases}$$

$
is $(2, 3)$ a solution to the system?
justify your answer.

Explanation:

Sub - Question 1: Solve the system of equations \(y = 2x-1\) and \(y=-x + 5\) by using a table.

For Table A (\(y = 2x-1\)):

Step 1: When \(x=-2\)

Substitute \(x=-2\) into \(y = 2x-1\).
\(y=2\times(-2)-1=-4 - 1=-5\)

Step 2: When \(x = 0\)

Substitute \(x = 0\) into \(y = 2x-1\).
\(y=2\times0-1=0 - 1=-1\)

Step 3: When \(x = 1\)

Substitute \(x = 1\) into \(y = 2x-1\).
\(y=2\times1-1=2 - 1 = 1\)

Step 4: When \(x = 2\)

Substitute \(x = 2\) into \(y = 2x-1\).
\(y=2\times2-1=4 - 1=3\)

Step 5: When \(x = 4\)

Substitute \(x = 4\) into \(y = 2x-1\).
\(y=2\times4-1=8 - 1 = 7\)

So Table A (for \(y = 2x - 1\)):

\(x\)	\(y\)
\(0\)	\(-1\)
\(1\)	\(1\)
\(2\)	\(3\)
\(4\)	\(7\)

For Table B (\(y=-x + 5\)):

Step 1: When \(x=-3\)

Substitute \(x=-3\) into \(y=-x + 5\).
\(y=-(-3)+5=3 + 5=8\)

Step 2: When \(x=-1\)

Substitute \(x=-1\) into \(y=-x + 5\).
\(y=-(-1)+5=1 + 5=6\)

Step 3: When \(x = 0\)

Substitute \(x = 0\) into \(y=-x + 5\).
\(y=-0 + 5=5\)

Step 4: When \(x = 2\)

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

Step 5: When \(x = 3\)

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

So Table B (for \(y=-x + 5\)):

\(x\)	\(y\)
\(-1\)	\(6\)
\(0\)	\(5\)
\(2\)	\(3\)
\(3\)	\(2\)

To find the solution of the system, we look for the \(x\) and \(y\) values that are the same in both tables. From the tables, when \(x = 2\), \(y = 3\) for both equations. So the solution of the system \(

$$\begin{cases}y = 2x-1\\y=-x + 5\end{cases}$$

\) is \(x = 2,y = 3\) or \((2,3)\).

Sub - Question 2: Is \((1,10)\) a solution to the system \(

$$\begin{cases}y = 8x+2\\y = 7x + 3\end{cases}$$

\)?

Step 1: Check the first equation \(y = 8x+2\)

Substitute \(x = 1,y = 10\) into \(y = 8x+2\).
Left - hand side (LHS) \(=y = 10\), Right - hand side (RHS) \(=8\times1+2=8 + 2 = 10\). So \(10=10\), the point satisfies the first equation.

Step 2: Check the second equation \(y = 7x+3\)

Substitute \(x = 1,y = 10\) into \(y = 7x+3\).
LHS \(=y = 10\), RHS \(=7\times1+3=7 + 3 = 10\). So \(10 = 10\), the point satisfies the second equation.

Since \((1,10)\) satisfies both equations of the system, \((1,10)\) is a solution to the system.

Sub - Question 3: Is \((2,3)\) a solution to the system \(

$$\begin{cases}3x-2y=-17\\5x + y=13\end{cases}$$

\)?

Step 1: Check the first equation \(3x-2y=-17\)

Substitute \(x = 2,y = 3\) into \(3x-2y\).
\(3\times2-2\times3=6 - 6 = 0\). But the right - hand side of the first equation is \(-17\). Since \(0
eq-17\), the point \((2,3)\) does not satisfy the first equation.

Since \((2,3)\) does not satisfy the first equation of the system, \((2,3)\) is not a solution to the system.

Answer:

s:

1. The solution of the system \(

$$\begin{cases}y = 2x-1\\y=-x + 5\end{cases}$$

\) is \(\boldsymbol{(2,3)}\).

1. Yes, \((1,10)\) is a solution to the system \(

$$\begin{cases}y = 8x+2\\y = 7x + 3\end{cases}$$

\) (because it satisfies both equations).

1. No, \((2,3)\) is not a solution to the system \(

$$\begin{cases}3x-2y=-17\\5x + y=13\end{cases}$$

\) (because it does not satisfy the first equation).