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Q: How do you solve b equals 2a plus 11 and a plus b equals 5 by substitution?

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r + 11 + 8r = 29

a = 4

x = 5, y = -1

-2x+5 x=-3 --original equation-2(-3)+5 --use substitution property6+5 --solve.11 --final answer is 11.

x=4 x=0

(0,7)

q = 78

9y subtract 11 equals 7

v is -15... -15*2 = -30... -30+11= -19.

4x-y=2, thereforey=4x-2sub y into 3x+2y=6...3x+8x-4=611x=10x=10/11

7

r + 11 = 19 r + 11 - 11 = 19 - 11 r = 8

If you mean: y = 6x+11 and y = 2x+7 Then by substitution: x = -1 and y = 5

You subtract 11 from both sides and r equals negative 7. See... r+11=3 -11 -11 r=7

1/5+1/11 = 11/55+5/55 = 16/55

If: 7+t = 11 Then: t = 11-7 So: t = 4

6a + 5a = -11 Combining like terms: 11a = -11 Dividing by 11: a = -1

If: x+y = 8 and 2x-y = 5 Then by adding the equations together: 3x = 13 or x = 13/3 By substitution into the original equations: x = 13/3 and y = 11/3

11 So: x = 37

-2x+13y=11 13y=2x+11 y=2x/13 + 11/13

22

5x-3 = 11+12x 5x-12x = 11+3 -7x = 14 x = -2

12 + 10 + 11 equals 33

-x + 5y = 10 5y= 11 y = 11/5

If 2x + 5 = 11, then 11 - 5 = 2x 6 = 2x 3 = x