Hello!
To solve this problem, we will use a system of equations. We will have one number be x and the other y. We will use substitutions to solve for each variable.
x+y=9
x=2y-9
To solve for the two numbers, we need to solve the top equation. The second equation shows that x=2y-9. In the first equation, we can replace 2y-9 for x and solve.
2y-9+y=9
3y-9=9
3y=18
y=6
We now know the value of y. Now we need to find x. We can plug in 6 for y in the second equation to find x.
x=2·6-9
x=12-9
x=3
Just to check, we will plug these two numbers into the first equation.
3+6=9
9=9
Our two numbers are three and six.
I hope this helps!
I suggest you using for those problems!
Answer:
B. 9/26.
Step-by-step explanation:
You need to change the coefficient of x in the first equation to 9/13 so that adding the 2 equations would eliminate x.
So you would multiply by
= 9/13 / 2
= 9/13 * 1/2
= 9/26 (answer)
They will be 12 miles apart a 2:30 pm, Trevor runs at a speed of 5 miles per hour, which they both start at noon which means, roughly in 12 miles it will be 2:30pm.
I'm assuming that you meant:
1
f(x) = -------- and that you want to find the value of x at which f(x) = h(x).
x+1
Of course you could create a table for each f(x) and h(x), but setting f(x)=h(x) and solving for x algebraically would be faster and more efficient:
1
f(x) = -------- = 2x + 3 = h(x). Then 1 = (x+1)(2x+3) = 2x^2 + 3x + 2x + 3
x+1
or 1 = 2x^2 + 5x + 3, or 2x^2 + 5x + 2 = 0.
This is a quadratic equation with a=2, b=5 and c=2. The discriminant is b^2-4ac, or 5^2-4(2)(2), OR 25-16= 9.
Thus, the roots are
-5 plus or minus sqrt(9)
x = ------------------------------------
2(2)
-5 plus or minus 3
= ----------------------------------
4
= {-1/2, -2}
Thus, f(x) = h(x) at both x=-1/2 and x= -2.