-3x+5y=-4
5y=3x-4
Y=3/5x-4/5
When it parallel to the line y=3/5x-4/5
It can be y=-3/5x+b
-5=6/5+b
B=-5-6/5=-25/5-6/5=-31/5
Y=-3/5x-31/5
3/5x+y=-31/5
Multiply 5
3x+5y=-31
In short, to convert two fractions to have the same denominator, we simply multiply one by the denominator of the other, so in this case, we'll multiply 1/3 by 5, top and bottom, and 1/5 by 3, top and bottom, thus
![\bf a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} \qquad \qquad \sqrt[ m]{a^ n}\implies a^{\frac{ n}{ m}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20a%5E%7B%5Cfrac%7B%20n%7D%7B%20m%7D%7D%20%5Cimplies%20%20%5Csqrt%5B%20m%5D%7Ba%5E%20n%7D%20%0A%5Cqquad%20%5Cqquad%0A%5Csqrt%5B%20m%5D%7Ba%5E%20n%7D%5Cimplies%20a%5E%7B%5Cfrac%7B%20n%7D%7B%20m%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \cfrac{15x^{\frac{1}{3}}}{y^{\frac{1}{5}}}\qquad \begin{cases} \frac{1}{3}=\frac{1\cdot 5}{3\cdot 5}\\ \qquad \frac{5}{15}\\\\ \frac{1}{5}=\frac{1\cdot 3}{5\cdot 3}\\ \qquad \frac{3}{15} \end{cases}\implies \cfrac{15x^{\frac{5}{15}}}{y^{\frac{3}{15}}}\implies 15\cdot \cfrac{x^{\frac{5}{15}}}{y^{\frac{3}{15}}}\implies 15\cdot \cfrac{\sqrt[15]{x^5}}{\sqrt[15]{y^3}} \\\\\\ 15\sqrt[15]{\frac{x^5}{y^3}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B15x%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%7By%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B1%5Ccdot%205%7D%7B3%5Ccdot%205%7D%5C%5C%0A%5Cqquad%20%5Cfrac%7B5%7D%7B15%7D%5C%5C%5C%5C%0A%5Cfrac%7B1%7D%7B5%7D%3D%5Cfrac%7B1%5Ccdot%203%7D%7B5%5Ccdot%203%7D%5C%5C%0A%5Cqquad%20%5Cfrac%7B3%7D%7B15%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B15x%5E%7B%5Cfrac%7B5%7D%7B15%7D%7D%7D%7By%5E%7B%5Cfrac%7B3%7D%7B15%7D%7D%7D%5Cimplies%2015%5Ccdot%20%5Ccfrac%7Bx%5E%7B%5Cfrac%7B5%7D%7B15%7D%7D%7D%7By%5E%7B%5Cfrac%7B3%7D%7B15%7D%7D%7D%5Cimplies%2015%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B15%5D%7Bx%5E5%7D%7D%7B%5Csqrt%5B15%5D%7By%5E3%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A15%5Csqrt%5B15%5D%7B%5Cfrac%7Bx%5E5%7D%7By%5E3%7D%7D)
Answer:
3y5 • (5y + 7)2 • (5y - 7) then solve that which is (3•2y5) • (5y + 7) • (5y - 7) and keep going { 3x−2y=5 & 2x−5y=7
SOLVE FOR X, Y
x=1
y=−1
I hope this helps a little
Answer:
The base of a power is the factor that is repeatedly multiplied by itself.
