Questions: Solve the system by graphing 6x + 4y = -12 x - 6y = -2

Transcript text: Solve the system by graphing \[ \begin{array}{c} 6 x+4 y=-12 \\ x-6 y=-2 \]

Solution

Solution Steps

Step 1: Write the equations in slope-intercept form

The given system of equations is:

$6x + 4y = -12$
$x - 6y = -2$

First, we will convert these equations into the slope-intercept form $y = mx + b$.

For the first equation: \[ 6x + 4y = -12 \] Subtract $6x$ from both sides: \[ 4y = -6x - 12 \] Divide by 4: \[ y = -\frac{3}{2}x - 3 \]

For the second equation: \[ x - 6y = -2 \] Subtract $x$ from both sides: \[ -6y = -x - 2 \] Divide by -6: \[ y = \frac{1}{6}x + \frac{1}{3} \]

Step 2: Identify the intersection point

To find the solution to the system, we need to find the intersection point of the two lines. We set the equations equal to each other:

\[ -\frac{3}{2}x - 3 = \frac{1}{6}x + \frac{1}{3} \]

Multiply every term by 6 to eliminate the fractions: \[ -9x - 18 = x + 2 \]

Add $9x$ to both sides: \[ -18 = 10x + 2 \]

Subtract 2 from both sides: \[ -20 = 10x \]

Divide by 10: \[ x = -2 \]

Substitute $x = -2$ back into one of the original equations to find $y$. Using the second equation: \[ y = \frac{1}{6}(-2) + \frac{1}{3} = -\frac{1}{3} + \frac{1}{3} = 0 \]

Final Answer

The solution to the system of equations is $x = -2$ and $y = 0$.

{"axisType": 3, "coordSystem": {"xmin": -5, "xmax": 5, "ymin": -5, "ymax": 5}, "commands": ["y = (-3/2)x - 3", "y = (1/6)x + (1/3)"], "latex_expressions": ["$y = -\\frac{3}{2}x - 3$", "$y = \\frac{1}{6}x + \\frac{1}{3}$"]}

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >