1)Solve the system by addition.?

Question:–4x – 5y = –13
x + 2y = 7
2)Solve the system by graphing.
x – y = 2
3x – 3y = 6

3)Graph f(x) = –2x + 2.

4)Graph f(x) = –x – 2.

5)The sum of two numbers is 31. Their difference is 7. What are the two numbers?

There's really no way of showing you the solution by graphing on EducationAsk.coms.

5) 12 and 19.
1) Write the two equations one on top of the other, matching the x, y and numeric factors. Pick a number to multiply each term of one equation so that either the x or y term has the same coefficient. Here, you can multiply the second equation by 4, making it 4x+8y=28. Add the two equations together and the x term is zero, so you can solve for y. Plug the y into the second equation, and you get the x value. Voila.

2) Unless you mistyped question 2, it's a trick question. The two equations describe the same line. So do 4x-4y-8 and 5x-5y=10, etc. So the two lines on a graph are the same, and there's no unique solution, which would be where the two lines intersect. There's no intersection here -- or maybe it's better to say the two lines intersect at every point.

3) and 4) are simple. Just choose two different values for x and get the y value (remember that on a graph, f(x) is the y value). In 3), if x is 2, y is -2. That's one point on the graph. If x is -2, y is 6. That's the second point. Draw a line between those points and you have the graph. Same thing in 4.

5) is a system of two equations. x+y=31 and x-y=7. Write them one above the other, add them and the y term drops out. Then you know what x is, and can find y easily.
1) Multiply bottom equation by 4 giving you 4x + 8y = 28. Add this to the top equation and the x terms cancel out leaving you with 3y = 15. Therefore, y = 5. Substitute this value into either equation to get the value of x. Into the bottome equation: x + 2(5) = 7. x = -3.

5) x + y = 31 and x - y = 7. Add equations and the y terms cancel leaving you with 2x = 38. x = 19. Again, substitute this value into either equation to get the y value of 12.

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