SOLUTION: Could you help me with this problem? 2x-y= -4
So far i have that you have to move the 2x over with the negative 4 and you subtract 2x from that and i don't know what to do afte
Algebra.Com
Question 138158This question is from textbook Algebra Structure and Method
: Could you help me with this problem? 2x-y= -4
So far i have that you have to move the 2x over with the negative 4 and you subtract 2x from that and i don't know what to do after that because you have to find the slope and the y-intercept and i don't know how to find it?
So could you please help me?
Thank you so much and i appreciate it!!!!!
This question is from textbook Algebra Structure and Method
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Do you have to find the slope and the y-intercept?
Rewrite the equation in the slope-intercept form: y = mx+b
2x-y = -4 Add y to both sides.
2x = y-4 Now add 4 to both sides.
2x+4 = y or
y = 2x+4 Compare with the slope-intercept form.
y = mx+b and you can see that the slope, m = 2 and the y-intercept, b = 4
RELATED QUESTIONS
Could you help me with the rest of this problem so far i have
2x > 8x - 24
-8x -8x (answered by rfer)
I am trying to solve an equation with fractions. This is the problem
2x over x^2+6x+8 (answered by stanbon)
Could I please have some help with solving this problem and also could you show me how... (answered by rfer)
Can you please help me with this problem the directions say **Multiply, then simplify... (answered by mikeg)
Im not sure if I chose the correct topic but could you please help me solve this... (answered by jim_thompson5910)
Could you please help me with this problem?
One number is twice another number. The... (answered by ankor@dixie-net.com)
Hi! Could you please help me with this problem?
I have to find the solutions for 2x^2... (answered by Alan3354)
I am having some trouble with quadratic formulas, could you please help me with this... (answered by ptaylor)
Could you please help me with the steps,in this kind of problem? I know that I have to... (answered by jim_thompson5910)