Find a parametric representation of the solution set of the linear equation. (enter your answer as a comma-separated list of equations. use s and t as your parameters.) −8x + 5y − 4z = 1

Step-by-step explanation:

There are at least 3 ways to do this problem, let us do it one way and then I'll explain how to do it from another approach. First of all let's begin by solving the linear equation for x. We get the following:

We can see that and   could act as free variables as we have no conditions or restrictions on the values of x and y. Normally one would see such restrictions by adding more equations as in a system of equations. That is why we can call   and   free variables.

If we for instance let   and   where   then we can write the linear equation as follows:

and are calle parametric variables.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )