Question

check your answers by graphing them.
1）\

$$\begin{cases} 2x - 3y = 12 \\\\ 2x - y = 0 \\end{cases}$$

sol

Explanation:

Step1: Solve the system of equations

We have the system:

$$\begin{cases} 2x - 3y = 12 \quad (1)\\ 2x - y = 0 \quad (2) \end{cases}$$

Subtract equation (2) from equation (1) to eliminate \(x\):
$$(2x - 3y)-(2x - y)=12 - 0$$
$$2x - 3y - 2x + y = 12$$
$$-2y = 12$$
Divide both sides by \(-2\):
$$y=\frac{12}{-2}=- 6$$

Step2: Substitute \(y\) into equation (2)

Substitute \(y = - 6\) into \(2x-y = 0\):
$$2x-(-6)=0$$
$$2x + 6 = 0$$
Subtract 6 from both sides:
$$2x=-6$$
Divide both sides by 2:
$$x=\frac{-6}{2}=-3$$

Answer:

The solution to the system of equations is \(x = - 3\), \(y=-6\), so the solution point is \((-3,-6)\)

To graph the lines:

• For the line \(2x - 3y=12\), we can find two points. When \(x = 0\), \(-3y=12\), \(y=- 4\), so the point is \((0,-4)\). When \(y = 0\), \(2x=12\), \(x = 6\), so the point is \((6,0)\).
• For the line \(2x-y = 0\) (or \(y = 2x\)), when \(x = 0\), \(y = 0\); when \(x = 1\), \(y = 2\), so the points are \((0,0)\) and \((1,2)\).

Plot these points and draw the lines. The lines will intersect at the point \((-3,-6)\), which confirms our solution.