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3 years ago

How can I solve each system of liner equations by substitution?

Mathematics

1 answer:

natka813 [3]3 years ago

8 0

I hope you can read cursive and I hope this helps! (:

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2y/5-y/4=-3 what is the common denominator

babymother [125]

Answer:

Step-by-step explanation:

3 0

3 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

1 year ago

Adrian purchases 6 tubes of paint and 3 paintbrushes at a craft store. Each tube of paint costs $2.50 and each paintbrush costs

Airida [17]

Answer:

$14.50

Step-by-step explanation:

6 x $2.50 = $15.00

3 x $0.50 = $1.50

$15.00 + $1.50 = $16.50

$16.50 - $2.00 = $14.50

8 0

3 years ago

Read 2 more answers

Write 1 2 3 to order from shortes to longest

Drupady [299]

1 then, 2 finally, 3

7 0

3 years ago

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Help please!!

denpristay [2]

Answer:

1 - 3 = 2/ 6x +2=8x 8x-2 = 6x divided by 6x = 1x so there but if 1 to ur math is no solution then theres no solution but if there is a solution ur welcome! 5*

8 0

3 years ago