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