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Solve the simultaneous equation

a)
5x+3y=41
2x+3y=40

b)
x+7y=64
x+3y=28

HELPPP

User R Pasha
by
8.7k points

1 Answer

2 votes

Answer:

a) (0.333, 13.11)

b) (1, 9)

Explanation:

a)

5x+3y=41

2x+3y=40

[The steps are labelled so they can be referenced in the subsequent problems]

A. Rearrange one of the two equations so as to isolate the x or y to one side. I'll use 2x+3y=40:

2x+3y=40

2x = 40-3y

x = (40-3y)/2

B. Now use that expression of x in the other equation:

5x+3y=41

5((40-3y)/2)+3y=41

(200-15y)/2 +3y = 41

100 - 7.5y + 3y = 41

-4.5y = - 59

y = 13.11

C. Now use y=13.11 in either equation to find x:

2x+3y=40

2x+3*(13.11)=40

2x + 39.33 = 40

2x = 0.67

x = 0.333

D. Answer: The lines intersect at (0.333, 13.11)

b)

x+7y=64

x+3y=28

A.

x+7y=64

x=64-7y

B.

x+3y=28

(64-7y)+3y=28

64-4y = 28

-4y = -36

y = 9

C.

x=64-7y

x=64-7*9

x = 1

D. Answer: The lines intersect at (1, 9)

See the attached graph for proof of the points of intersection.

HELPPP PROVIDED (I hope)

Solve the simultaneous equation a) 5x+3y=41 2x+3y=40 b) x+7y=64 x+3y=28 HELPPP-example-1
User Bouna
by
8.0k points