Solve the system algebraically check your work 2x+y- 10=0; x-y-5=0

x = 5 and y = 0 is correct solution of 2x + y -10 = 0 and x – y – 5 =0

Solution:

Two given equations which needs to be solve are  

2x + y – 10 = 0    ------ (1)

x– y – 5 = 0       ------ (2)  

Let’s modify equation (1)

2x + y – 10 = 0    

y =10 - 2x        ------ (3)

On substituting value of y from equation (3) in equation (2) we get

x – (10 – 2x) -5 = 0

x – 10 + 2x – 5 = 0

3x -15 = 0

x = 5

Substituting x = 5 in equation (3) to get value of y.

y = 10 – 2 5 = 10 – 10 = 0

So on solving given equation we get x = 5 and y = 0.

Lets substitute value of x = 5 and y = 0 in equation (1) and equation (2) to check whether these calculated values satisfies given equations or not.

For equation (1), 2 5 + 0 – 10 = 10 – 10 = 0

For equation (2), 5 – 0 – 5 = 0  

On solving, in both cases LHS = RHS for calculated values of x = 5 and y = 0.

Hence x = 5 and y = 0 is correct solution of two given equation.

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

english maybe i can ..