Question

Which ordered pair is a solution to the system of linear equations One-half x minus three-fourths y = StartFraction 11 Over 60 EndFraction and Two-fifths x + one-sixth y = StartFraction 3 Over 10 EndFraction? (one-fifth, negative two-thirds) (one-fifth, two-thirds) (two-thirds, negative one-fifth) (two-thirds, one-fifth)

Answer 1

Answer:

its b

Step-by-step explanation:

just took the test

Answer 2

Answer:

$\left(\dfrac{2}{3},\dfrac{1}{5}\right)$

Step-by-step explanation:

Given the system of two equations:

$\left\{\begin{array}{l}\dfrac{1}{2}x-\dfrac{3}{4}y=\dfrac{11}{60}\\ \\\dfrac{2}{5}x+\dfrac{1}{6}y=\dfrac{3}{10}\end{array}\right.$

Multiply the first equation and the second equation by 60 to get rid of fractions:

$\left\{\begin{array}{l}30x-45y=11\\ \\24x+10y=18\end{array}\right.$

Now multiply the first equation by 4 and the second equation by 5:

$\left\{\begin{array}{l}120x-180y=44\\ \\120x+50y=90\end{array}\right.$

Subtract them:

$(120x-180y)-(120x+50y)=44-90\\ \\120x-180y-120x-50y=-46\\ \\-230y=-46\\ \\y=\dfrac{46}{230}=\dfrac{1}{5}$

Substitute it into the first equation:

$30x-45\cdot \dfrac{1}{5}=11\\ \\30x-9=11\\ \\30x=11+9\\ \\30x=20\\ \\x=\dfrac{2}{3}$

The solution is $\left(\dfrac{2}{3},\dfrac{1}{5}\right)$