The solution of the system of equations is (6 , -7)
Step-by-step explanation:
There are two method to solve the system of equations
- Elimination method: we make the coefficients of one variable in the two equations have same values and different signs, then we add the two equations to eliminate this variable and have an equation of other variable, we solve it to find the other variable, then substitute the value of this variable in one of the two equations to find the first variable
- Substitution method: We use one of the two equations to find one variable in terms of the other, then substitute it in the second equation to have an equation of the other variable, we solve it to find the other variable, then substitute the value of this variable in the equation of the first variable
Let us use the elimination method with your problem
∵ 3x + 2y = 4 ⇒ (1)
∵ 3x + 6y = -24 ⇒ (2)
- Multiply equation (1) by -1 to eliminate x ⇒ to make the coefficients of x in the two equations have same values and different signs
∵ -3x - 2y = -4 ⇒ (3)
- Add equations (2) and (3)
∴ 4y = -28
- Divide both sides by 4
∴ y = -7
Substitute value of y in equations (1) OR (2) to find x
∵ 3x + 2(-7) = 4
∴ 3x - 14 = 4
- Add 14 for both sides
∴ 3x = 18
- Divide both sides by 3
∴ x = 6
The solution of the system of equations is (6 , -7)
<em>I hope this explanation help you</em>
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I believe it would be A. Political thinking. Hope it helps! :)
The answer is you could by 6 games. mostly because you would have 10 dollors left . so now you can add me as a friend my name is angel1600 yessss.
now, let's recall the rational root test, check your textbook on it.
so p = 18 and q = 1
so all possible roots will come from the factors of ±p/q
now, to make it a bit short, the factors are loosely, ±3, ±2, ±9, ±1, ±6.
recall that, a root will give us a remainder of 0.
let us use +3.
![\bf x^4-7x^3+13x^2+3x-18 \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{r|rrrrr} 3&1&-7&13&3&18\\ &&3&-12&3&18\\ \cline{1-6} &1&-4&1&6&0 \end{array}\qquad \implies (x-3)(x^3-4x^2+x+6)](https://tex.z-dn.net/?f=%5Cbf%20x%5E4-7x%5E3%2B13x%5E2%2B3x-18%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Br%7Crrrrr%7D%203%261%26-7%2613%263%2618%5C%5C%20%26%263%26-12%263%2618%5C%5C%20%5Ccline%7B1-6%7D%20%261%26-4%261%266%260%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%28x-3%29%28x%5E3-4x%5E2%2Bx%2B6%29)
well, that one worked... now, using the rational root test, our p = 6, q = 1.
so the factors from ±p/q are ±3, ±2, ±1
let's use 3 again
and of course, we can factor x²-x-2 to (x-2)(x+1).
(x-3)(x-3)(x-2)(x+1).
Answer:
8/15
Step-by-step explanation:
