Question

3 part 2
solving systems of equations
unit 3 part 2 test: b
solve the quadratic equation by taking square roots.

1. given the graph of the system of equations find the

solution the system.

$$ left{ \begin{array} { l } { y = - \frac { 2 } { 3 } x + 8 } \\ { y = \frac { 1 } { 2 } x + 1 } end{array} ight. $$

solution:

1. classify the following system and determine the number of solutions and explain your

answer.

$$ left{ \begin{array} { l } { y = - 2 x + 10 } \\ { y = - 2 x + 6 } end{array} ight. $$

circle the correct classification
consistent dependent consistent independent inconsistent
the system of equation has _ solution(s) because _

1. solve the following linear system by graphing

$$ left{ \begin{array} { l } { y = - x + 3 } \\ { y = x + 1 } end{array} ight. $$

Explanation:

Step1: Solve the first system

Set \(-\frac{2}{3}x + 8=\frac{1}{2}x + 1\).
Multiply through by 6 to clear fractions: \(-4x+48 = 3x+6\).
Add \(4x\) to both sides: \(48=7x + 6\).
Subtract 6: \(42 = 7x\).
Divide by 7: \(x = 6\).
Substitute \(x = 6\) into \(y=\frac{1}{2}x + 1\): \(y=\frac{1}{2}(6)+1=3 + 1=4\).

Step2: Analyze the second system

The equations \(y=-2x + 10\) and \(y=-2x+6\) have the same slope (\(m=-2\)) but different \(y\) - intercepts (\(b = 10\) and \(b = 6\)). Parallel lines never intersect.

Step3: Solve the third system

Set \(-x + 3=x + 1\).
Add \(x\) to both sides: \(3=2x + 1\).
Subtract 1: \(2=2x\).
Divide by 2: \(x = 1\).
Substitute \(x = 1\) into \(y=x + 1\): \(y=1 + 1=2\).

Answer:

1. \((6,4)\)
2. Circle “Inconsistent”. The system of equation has \(0\) solution(s) because the lines are parallel (same slope, different \(y\) - intercepts)
3. \((1,2)\)