Consider the following system with a solution (1,3) equation 1 of the system 2x+y=5 equation 2 of the system x-2y=-5 prove that replacing the first equation with the sum of the equation and a multiple of the other produces a system with the same solution

The prove in the procedure

Step-by-step explanation:

we have

2x+y=5 > equation 1

x-2y=-5 > equation 2

so

Replace the first equation with the sum of the equation and a multiple of the other

Multiply equation 2 by 3 both sides

3*(x-2y)=-5*3

3x-6y=-15 > equation 3

Adds equation 3 and equation 1

3x-6y=-15

2x+y=5

3x+2x-6y+y=-15+5

5x-5y=-10 > equation 4

The new system is

5x-5y=-10 > equation 4

x-2y=-5 > equation 2

Solve the system by elimination

Multiply equation 2 by -5 both sides

-5*(x-2y)=-5*(-5)

-5x+10y=25 > equation 5

Adds equation 4 and equation 5

5x-5y=-10

-5x+10y=25

-5y+10y=-10+25

5y=15

y=3

Find the value of x

substitute the value of y in the equation 2

x-2y=-5

x-2(3)=-5

x-6=-5

x=-5+6

x=1

The solution of the new system of equations is (1,3)

therefore

The solution of the new system of equations is the same solution of the original system of equations.

42%

step-by-step explanation:

we know,

total people = 60

total males = 25

percentage of males

= 25/60 × 100

= 25/6 × 10

= 25/3 × 5

= 125 / 3

= 41.

rounding of = 42%

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