plz help me understand this
1.) 6x - 3y = 6 2.) y= -2/3x +4
y= 2x + 5 2x + 3y = -6
3.) x+y=0 4.) 3x - 4y = -5
x=y+4 x = y+2
What are we supposed to do
I see a bunch of equations running together ....
1.) 6x - 3y = 6 ---> y = 2x - 2
etc ?
I think I see systems of equations:
1.) 6x - 3y = 6
y= 2x + 5
2.) y= -2/3x +4
2x + 3y = -6
3.) x+y=0
x=y+4
4.) 3x - 4y = -5
x = y+2
All you need to do here is set up things in a standard way and solve the pairs of equations. I'll do #2 and you can try the others.
#2.
y= -2/3x +4
2x + 3y = -6
2/3 x + y = 4
2x + 3y = -6
2x + 3y = 12
2x + 3y = -6
Hmm. no solution here
Let's try #1:
6x - 3y = 6
y= 2x + 5
6x - 3y = 6
-2x + y = 5
multiply the 2nd by 3 and add:
6x - 3y = 6
-6x + 3y = 15
WAK! Again no solution.
#3.
x+y=0
x=y+4
x + y = 0
x - y = 4
add 'em
2x = 4
x = 2
y = -2
#4.
3x - 4y = -5
x = y+2
3x - 4y = -5
x - y = 2
multiply the 2nd by 3 and subtract the 1st:
3x - 4y = -5
3x - 3y = 6
y = 11
x = 13
If I read the problems wrong, write them right and you take a stab at solving them.
Sure, I'd be happy to help you understand these equations. Let's go through each pair of equations step by step.
1.) 6x - 3y = 6 y = 2x + 5
To solve this system of equations, we can use the method of substitution. We can start by solving one equation for one variable and then substituting that expression into the other equation.
In the second equation, y is already isolated, so we can substitute it into the first equation:
6x - 3(2x + 5) = 6
Simplify the expression:
6x - 6x - 15 = 6
-15 = 6
This is not a true statement, which means the system of equations has no solution. The lines represented by these equations are parallel and will never intersect.
2.) y= 2x + 5 2x + 3y = -6
For this system of equations, we can use the method of substitution or the method of elimination.
Let's use the method of substitution. Since y is already isolated in the first equation, we can substitute it into the second equation:
2x + 3(2x + 5) = -6
Simplify the expression:
2x + 6x + 15 = -6
8x + 15 = -6
Subtract 15 from both sides:
8x = -21
Divide both sides by 8:
x = -21/8
Now we can substitute this value of x back into the first equation to solve for y:
y = 2(-21/8) + 5
y = -21/4 + 20/4
y = -1/4
So the solution to this system of equations is x = -21/8 and y = -1/4.
3.) x+y=0 x = y+4
In this system of equations, we can solve it using the method of substitution or the method of elimination.
Let's use the method of substitution. We can solve the second equation for x in terms of y:
x = y + 4
Now we can substitute this expression for x into the first equation:
(y + 4) + y = 0
2y + 4 = 0
Subtract 4 from both sides:
2y = -4
Divide both sides by 2:
y = -2
Now, substitute this value of y back into the second equation to solve for x:
x = -2 + 4
x = 2
So the solution to this system of equations is x = 2 and y = -2.
4.) 3x - 4y = -5 x = y + 2
For this system of equations, we can use the method of substitution or the method of elimination.
Let's use the method of substitution. We can solve the second equation for x in terms of y:
x = y + 2
Now substitute this expression for x into the first equation:
3(y + 2) - 4y = -5
Simplify the expression:
3y + 6 - 4y = -5
-y + 6 = -5
Subtract 6 from both sides:
-y = -11
Multiply both sides by -1 to isolate y:
y = 11
Now substitute this value of y back into the second equation to solve for x:
x = 11 + 2
x = 13
So the solution to this system of equations is x = 13 and y = 11.
I hope this helps you understand how to solve these systems of equations! Let me know if you have any other questions.