The solution is the point of intersection between the two equations.
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
I wouldn’t change anything because that is how I lived my life
1 x 64=64
2 x 32=64
4 x 16 =64
8x8=64
Answer:
9 1/6 yd
Step-by-step explanation:
First, list the lengths of ribbon Laura had or acquired:
8 3/10 yd + 4 3/5 yd. This is equivalent to 8 3/10 yd + 4 6/10 yd, or
12 9/10 yd.
Next, list the lengths of ribbon Laura used:
1 2/5 yd + 2 1/3 yd. The LCD here is 15, so we have 1 6/15 yd + 2 5/15 yd.
Summing up, we get 3 11/15 yd.
Subtract this length from the original 12 9/10 yard with which Laura began:
12 9/10 yd or 12 27/30 yd (LCD is 30)
- 3 11/15 yd or -3 22/30 yd
Subtracting the latter two quantities, we get 9 5/30 yd, or 9 1/6 yd
Six women push grocery carts up a ramp as shown.
2 carts weigh 50 lbs. each.
4 carts weigh 20 lbs. each.
How much work was done?
_____________ ft.-lbs. (total work)