5x+2y=6 3y=2x+9 how many solutions does this problem have? No solution, one solution, or infinite solutions

Mathematics

1 answer:

Komok [63]2 years ago

8 0

Answer:

One solution

Step-by-step explanation:

First, organize the equations by moving y to the left hand side:

5x + 2y = 6, subtract 5x from both sides and we get: 2y=6-5x

3y = 2x + 9 is already done

Next, set the coeficcient of both y values equal to the least common multiple of 2 and 3 which is 6:

2y=6-5x multiply both sides by 3, 6y = 18-15x

3y=2x+9 multiply both sides by 2, 6y = 4x+18

Substitute y on both sides and we get 18-15x=4x+18

Simplify that and you get 19x=0, x equals 0, so there is one solution

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Which expression is equivalent to square root 55x^7y6/11x^11y^8

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$\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}$

Step-by-step explanation:

$\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a$

$=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}$