Solve the solution of equations. 3x + 4y = -23 X = 3y + 1 X = Y =

Mathematics

2 answers:

Lena [83]3 years ago

8 0

Answer:

x = -5

y = -2

Step-by-step explanation:

Greeley [361]3 years ago

4 0

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}3x+4y=-23&(1)\\x=3y+1&(2)\end{array}\right\\\\\text{Substitute (2) to (1):}\\\\3(3y+1)+4y=-23\qquad\text{use the distributive property}\\\\(3)(3y)+(3)(1)+4y=-23\\\\9y+3+4y=-23\qquad\text{subtract 3 from both sides}\\\\13y=-26\qquad\text{divide both sides by 13}\\\\\boxed{y=-2}\\\\\text{Put the value of }\ y\ \text{to (2)}:\\\\x=3(-2)+1\\\\x=-6+1\\\\\boxed{x=-5}$

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Can someone please help me?

Nookie1986 [14]

Answer:

-3 • (x - 7)

Step 1 :

6x - 3 • (3x - 7)

Step 2:

21 - 3x = -3 • (x - 7)

Final result :

-3 • (x - 7)

8 0

4 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

$(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .