For time t≥0 , the acceleration of a particle moving along the x-axis is given by a(t)=cost+t. At time t=0, the velocity of the particle is 8 and the position of the particle is 0. What is the position of the particle at time t=π4 ?

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7  -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so   14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as  -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84  then x = - 0.92 or around -0.9best luck!

Ci think but i'm not entirely sure so..