Solve the system of linear equations below.

2x + y = 5
x + y = 4
a.
x = 1, y = 3
b.
x = 1, y = 5
c.
x = 3, y = 1
d.
x = 9, y = -5

x = 1 , y = 3 thus: A is your Anser

Step-by-step explanation:

Solve the following system:

{2 x + y = 5 | (equation 1)

x + y = 4 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y = 5 | (equation 1)

0 x+y/2 = 3/2 | (equation 2)

Multiply equation 2 by 2:

{2 x + y = 5 | (equation 1)

0 x+y = 3 | (equation 2)

Subtract equation 2 from equation 1:

{2 x+0 y = 2 | (equation 1)

0 x+y = 3 | (equation 2)

Divide equation 1 by 2:

{x+0 y = 1 | (equation 1)

0 x+y = 3 | (equation 2)

Collect results:

 {x = 1 , y = 3

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

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step-by-step explanation: