Answer:
(-4,3)
Step-by-step explanation:
To solve a system of equations by graphing, we'll need to graph both equations, and find their points of intersection.
Note that both equations are linear (no exponents, no radicals, no variables in a denominator, no variables multiplied to other variables, etc -- just numbers multiplied to a variable and added to other numbers multiplied to a variable).
To graph linear equations, often they are graphed by putting the equation in slope-intercept form. Alternatively, since it is a line, two points can be found on the line, and then the line can be graphed.
Option 1: Convert to slope-intercept form
Slope intercept form is [tex]y=mx+b[/tex], where "m" is the slope of the line, and "b" is the y-intercept (the place where the line crosses the y-axis).
To convert to slope intercept form, isolate "y".
First equation:
[tex]4y+3x=0[/tex]
[tex](4y+3x)-3x=(0)-3x[/tex]
[tex]4y=-3x[/tex]
[tex]\dfrac{4y}{4}=\dfrac{-3x}{4}[/tex]
[tex]y=\frac{-3}{4}x[/tex]
Second equation:
[tex]4y-x=16[/tex]
[tex](4y-x)+x=(16)+x[/tex]
[tex]4y=x+16[/tex]
[tex]\frac{1}{4}*(4y)=\frac{1}{4}*(x+16)[/tex]
[tex]y=\frac{1}{4}*x+\frac{1}{4}*16[/tex]
[tex]y=\frac{1}{4}x+4[/tex]
To graph the lines, plot their y-intercepts first, then use their slopes to determine the rest of the line.
Recall that the slope is [tex]\frac{rise}{run}[/tex].
Once the equations are graphed, find the intersection from the diagram, (-4,3).
Option 2: Graphing from implicit form
The equations currently are in an implicit form (a form where the variables aren't isolated, so neither variable is written in terms of the other). To graph any line, find and plot two points, then draw the line between them.
To find points on the line, recall that the equation for a line relates the x-coordinate and y-coordinate through the equation. So, if you want to find the y-coordinate for the line when the x-coordinate is 0, substitute 0 for x, and solve for y. Often zero is used, because multiplying by zero cancel out the term, and makes the calculations easier.
Equation 1 - finding a point where x=0
[tex]4y+3x=0[/tex]
[tex]4y+3(0)=0[/tex]
[tex]4y+0=0[/tex]
[tex]4y=0[/tex]
[tex]\dfrac{4y}{4}=\dfrac{0}{4}[/tex]
[tex]y=0[/tex]
So, if x=0, then y=0. So, the point (0,0) is on line 1.
To find another point, we need to choose another number. As mentioned previously, often zero is used, and we could find a point on the line where the y-coordinate is zero. However, since we just found that the point (0,0) is on the line, x=0 when y=0, and so y=0 when x=0. We'll need a new number.
Another number that it often an easy choice mathematically is to choose the coefficient of the other variable. So, for instance, in equation 1, the coefficient of the y term is "4", so let's choose the x-coordinate to be 4, and find the y-coordinate that goes with it:
Equation 1 - finding a second point, where x=4
[tex]4y+3x=0[/tex]
[tex]4y+3(4)=0[/tex]
[tex]4y+12=0[/tex]
[tex](4y+12)-12=(0)-12[/tex]
[tex]4y=-12[/tex]
[tex]\dfrac{4y}{4}=\dfrac{-12}{4}[/tex]
[tex]y=-3[/tex]
So, if x=4, then y=-3. So, the point (4,-3) is also on line 1.
Those two points can be plotted, and the line drawn.
Equation 2 - finding a point where x=0
[tex]4y-x=16[/tex]
[tex]4y-(0)=16[/tex]
[tex]4y=16[/tex]
[tex]\dfrac{4y}{4}=\dfrac{16}{4}[/tex]
[tex]y=4[/tex]
So, if x=0, then y=4. So, the point (0,4) is on line 2.
To find another point, this time, we can choose the y-coordinate to be zero, because we don't already know the x-coordinate that is associated with it.
Equation 2 - finding a second point, where y=0
[tex]4y-x=16[/tex]
[tex]4(0)-x=16[/tex]
[tex]0-x=16[/tex]
[tex]-x=16[/tex]
[tex]-1*(-x)=-1*(16)[/tex]
[tex]x=-16[/tex]
So, if y = 0, then x = -16. So, the point (-16,0) is also on line 2.
Those two points can be plotted and the line drawn. Once the lines are drawn, the intersection can be found. From the diagram, the intersection is (-4,3).
