a+b=10 and 3a+2b=40?

Question

a+b=10 and 3a+2b=40

Answer 1

a + b = 10 ...(1)

3a + 2b = 40 ...(2)



multiply (1) with 2 at bothside

2a + 2b = 20 ...(3)



Subtract (3) from (2)

a = 20



a + b = 10

20 + b = 10

b = -10



a = 20 , b = -10

Answer 2

A = 20, B = -10

Answer 3

3a + 2b = 40

3a = 40 - 2b

a = (40 - 2b)/3

((40-2b)/3) + b = 10

((40-2b)/3) + 3b/3 = 10

((40-2b+3b)/3) = 10

(40+b)/3 = 10

(3 * (40 + b) / 3) = 3 * 10

40 + b = 30

b = -10



3a + 2b = 40

3a + 2(-10) = 40

3a - 20 = 40

3a = 60

a = 20



a = 20

b = -10



a + b = 10

20 + -10 = 10

20 - 10 = 10

Answer 4

a = 20, b = -10

Answer 5

a+b = 10

a = 10-b --------eq 1

3a+2b = 40 ---eq2





substitute eq1 in eq2

3(10-b)+2b = 40

30-3b+2b = 40

30-b=40

-b = 40-30=10

-b=10 multiply by -1

b=-10

subistitute in eq1

a=10-b

a=10-(-10)

a=10+10=20

best of luck!!!!

Answer 6

its either multiply a+b=10 by 3 to eliminate a, or multiply a+b=10 by 2 to eliminate b

lets use 2, so

2(a+b=10) get 2a+2b=20



2a+2b=20 minus

3a+2b=40

get -a=-20



therefore a =20, then substitute a =20 to either of the 2 original equation



a+b=10

20+b=10

b=10-20

b=-10

Answer 7

a+b=10 and 3a+2b=40



SOLUTION:



a+b=10 eq.1

3a+2b=40 eq.2



from eq.1,

a+b=10

a = 10 - b eq.3



solve for b, in terms of a.used eq.2 and eq.3



3a+2b=40 eq.2

a = 10 - b eq.3



substitute the value of a,



3 (10 - b) +2b = 40



simplify this



30 - 3b + 2b = 40



30 - 40 = 3b - 2b



b = -10,



you the value of b, u substitute in eq1 or eq2 to solve for a,



a+b=10 eq.1



a - 10 = 10

a =10 +10

a = 20



hope it helps,

Answer 8

a+b=10----------(1)

3a+2b=40-------(2)

(1)x3-(2)x1---->b= - 10

substituting this value of b in (1)

we get a=20

If a=20,b= -10 and a+b=10 then 3a+2b=40.

Answer 9

another way, rather than elimination is by substitution..

so...



a+b=10 becomes a=10-b (1)

substitute a from equation 1 to the 2nd equation... so...



3(10-b)+2b=40

30-3b+2b=40

-b=10

b=-10



now, substitute b to the first equation



a+(-10)=10

a-10=10

a=10+10

a=20



:))

Answer 10

simply by using the elimination method

looking at the first and second equation we found out that if we multiply the 1st eq by negative two we will come up by eliminating the variable b in the second eq.



ist eq

(a + b)x-2=10x-2

thus

-2a-2b=-20



so:



-2a-2b=-20

3a+2b=40



thus we eliminate b



so a=20



substituting to the first eq.



20 +b = 10



b now equals -10



to check if your answer is correct simply substitute the values that you get in any of the two eq and chek if the left and right side of the eq is correct



a + b = -10



10 - 20 = -10

Answer 11

a=20..b=-10

Answer 12

a+b=10..so, a=10-b

therefore, 3a+2b=40

3(10-b)+2b=40

30-3b+2b=40

30-b=40

-b=10

b=-10



we kno that

a=10-b

a=10-(-10)

a=20



so a=20 & b=-10

Answer 13

to sparkling up a device by applying substituting, you may desire to set up between the equations so as that a variable is remoted. To isolate a variable interior the 1st equation, subtract one among them. the two b=10+a and a=10-b are achieveable, even though it relies upon on what you're finding to sparkling up. in case you prefer to sparkling up for a, then use the 1st. sparkling up for b, use the 2nd. desire this helped!

Answer 14

a+b=10

3a+2b=40



a=10-b



3(10-b)+2b=40

30-3b+2b=40

30-b=40

-b=10

b=-10



a-10=10

a-10+10=10+10

a=20



a=20 b=-10

Answer 15

a = 20

b = -10

Answer 16

simple algebra equation

a=20

b=10.

that very easy

Answer 17

a+b=10

3a+2b=40

a+2a+2b=40

a+2(a+b)=40

a+20=40

a=20

b=10-20

b=-10