X*a = 244 is equation (1)
x+a = 2 is equation (2)
Solve equation (2) for 'a' to get
x+a = 2
a = 2-x
Call this equation (3)
We will plug equation (3) into equation (1)
x*a = 244
x*(a) = 244
x*(2-x) = 244
Notice how the 'a' is replaced with an expression in terms of x
Let's solve for x
x*(2-x) = 244
2x-x^2 = 244
x^2-2x+244 = 0
If we use the discriminant formula, d = b^2 - 4ac, then we find that...
d = b^2 - 4ac
d = (-2)^2 - 4*1*244
d = -972
indicating that there are no real number solutions to the equation x^2-2x+244 = 0
So this means that 'x' and 'a' in those two original equations are non real numbers. If you haven't learned about complex numbers yet, then the answer is simply "no solution". At this point you would stop here.
If you have learned about complex numbers, then the solution set is approximately
{x = 1 + 15.588i, a = 1 - 15.588i}
which can be found through the quadratic formula
Note: it's possible that there's a typo somewhere in the problem that your teacher gave you.
Answer: SU = 4(1) + 1 = 5
Step-by-step explanation:
Since T is on segment SU we know the whole is eqaul to the sum of it’s parts.
ST + TU = SU substitute
3x - 1 + 3x = 4x + 1 simplify
6x - 1 = 4x + 1 solve for x
2x = 2
x = 1
ST = 3(1) -1 = 2
TU = 3(1) = 3
SU = 4(1) + 1 = 5
Answer:
Step-by-step explanation:
<u>First, Let's solve for x in -2x+2y=-4:</u>
<u>Subtract 2y from both sides:</u>
<u>Divide both sides by -2:</u>
<u>Now, we'll substitute x=y+2 to 3x+3y=-18:</u>
→ let x=2+y
<u>Simplify:</u>
<u>Now, let's solve for y in 6+6y=-18</u>
<u>Subtract 6 from both sides:</u>
<u>Divide both sides by 6:</u>
<u>Now, substitute y=-4 into x=2+y:</u>
→ let y = -4
Therefore, x=-2 and y=-4.
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