Question

graph the systems of linear equations to find the solutions.
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$$\begin{cases} x + 2y = 6 \\\\ x - y = 3 \\end{cases}$$

Explanation:

Step1: Rewrite equations in slope - intercept form

For the first equation \(x + 2y=6\), solve for \(y\):
Subtract \(x\) from both sides: \(2y=-x + 6\).
Divide by 2: \(y=-\frac{1}{2}x + 3\).
For the second equation \(x - y=3\), solve for \(y\):
Subtract \(x\) from both sides: \(-y=-x + 3\).
Multiply by - 1: \(y=x - 3\).

Step2: Find intercepts for first line (\(y =-\frac{1}{2}x+3\))

• \(y\) - intercept: When \(x = 0\), \(y=3\). So the point is \((0,3)\).
• \(x\) - intercept: When \(y = 0\), \(0=-\frac{1}{2}x + 3\), \(\frac{1}{2}x=3\), \(x = 6\). So the point is \((6,0)\).

Step3: Find intercepts for second line (\(y=x - 3\))

• \(y\) - intercept: When \(x = 0\), \(y=-3\). So the point is \((0,-3)\).
• \(x\) - intercept: When \(y = 0\), \(0=x - 3\), \(x = 3\). So the point is \((3,0)\).

Step4: Graph the lines

• For \(y =-\frac{1}{2}x + 3\), plot \((0,3)\) and \((6,0)\) and draw the line.
• For \(y=x - 3\), plot \((0,-3)\) and \((3,0)\) and draw the line.

Step5: Find the intersection point

The two lines intersect where \(y=-\frac{1}{2}x + 3\) and \(y=x - 3\) are equal.
Set \(-\frac{1}{2}x + 3=x - 3\).
Add \(\frac{1}{2}x\) to both sides: \(3=\frac{3}{2}x-3\).
Add 3 to both sides: \(6=\frac{3}{2}x\).
Multiply by \(\frac{2}{3}\): \(x = 4\).
Substitute \(x = 4\) into \(y=x - 3\), \(y=4 - 3=1\). So the intersection point is \((4,1)\).

Answer:

The solution of the system of linear equations is \(x = 4\), \(y = 1\) (the point of intersection of the two lines is \((4,1)\)).