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2 years ago

Solve using substitution method x-y=11;3x+10y=-6 explain please

1 answer:

6 0

Ok so for starters you want to choose an equation and solve for a variable.
So, I am going to choose x from the first equation.
Add y to both sides and you get x=11+y
Next, substitute 11+y for x in the other equation so you get...
2(11+y) +10y=-6
Next distribute the 2 throug the 11 and the y
22+2y+10y=-6
12y=-28
y=-28/12
reduce this fraction to make this easier.
y=-7/3
Now plug in why to either of the equations to find x
x-(-7/3)=11
x+7/3=11
x=11-(7/3)
x=(33/3)-(7/3)
x=26/3
so x = 26/3 and y = -7/3
you can also check to see if this is correct by substituting each of these values into the equations.

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Determine the intercepts of the line

Elanso [62]

Answer:

y- (0,2.5) x-(3.5,0)

Step-by-step explanation:

This is because the intercepts of a graph are the points at which a line intersects with a desired axis. So for the Y axis the inercept would be at 2.5 making the y intercept (0,2.5). The same concept is used on the X axis.

5 0

2 years ago

2/3=n/12 solve the proportion

Natali [406]

Answer: $n=8$

Cross-multiply

$2/3=n/12$

$(2)*(12)=n*(3)$

$24=3n$

Flip the equation

$24=3n---->3n=24$

Divide both sides by 3

$3n/3=24/3$

$24/3=8----> n=8$

5 0

3 years ago

Use the distributive property to write an expression equivalent to 5x (3+4)

Law Incorporation [45]

You can solve it algebraically by simply adding the numbers in the parenthesis and multiplying to get 35x but if you want to solve it using distributive property exclusively then distribute as so

5x(3) + 5x(4)

15x + 20x

35x

8 0

3 years ago

Read 2 more answers

Use the row operations tool to solve the following system of equations, obtaining the solutions in fraction form.

Inessa05 [86]

Answer:

$x=-\frac{7}{26}, y=\frac{37}{26}, z=-\frac{21}{13}$

Step-by-step explanation:

The matrix representation of the system of linear equations is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\11&13&9&\vdots&1\\3&5&7&\vdots&-5\\8&8&2&\vdots&6\\7&5&2&\vdots&2\end{array}\right)$

First apply the following row operations:

$\begin{array}{c}R_{2}\to R_{2}+(-\frac{11}{15})R_{1}\\R_{3}\to R_{3}+(-\frac{1}{5})R_{1}\\R_{4}\to R_{4}+(-\frac{8}{15})R_{1}\\R_{5}\to R_{5}+(-\frac{7}{15})R_{1}\end{array}$

The resulting matrix is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&\frac{52}{15}&\frac{91}{15}&\vdots&-\frac{73}{15}\\0&\frac{12}{5}&\frac{31}{5}&\vdots&-\frac{33}{5}\\0&\frac{16}{15}&-\frac{2}{15}&\vdots&\frac{26}{15}\\0&-\frac{16}{15}&\frac{2}{15}&\vdots&-\frac{26}{15}\end{array}\right)$

Then apply the row operations:

$\begin{array}{c}R_{3}\to R_{3}+(-\frac{12}{5}\cdot \frac{15}{52})R_{2}\\R_{4}\to R_{4}+(-\frac{-16}{15}\cdot \frac{15}{52})R_{2}\\R_{5}\to R_{5}+(\frac{16}{15}\cdot \frac{15}{52})R_{2}\end{array}$

The resulting matrix is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&\frac{52}{15}&\frac{91}{15}&\vdots&-\frac{73}{15}\\0&0&2&\vdots&-\frac{42}{13}\\0&0&-2&\vdots&\frac{42}{13}\\0&0&2&\vdots&-\frac{42}{13}\end{array}\right)$

Now apply the row operations:

$\begin{array}{c} R_{4}\to R_{4}+R_{3}\\R_{5}\to R_{5}+(-1)R_{3}\end{array}$

The resulting matrix is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&\frac{52}{15}&\frac{91}{15}&\vdots&-\frac{73}{15}\\0&0&2&\vdots&-\frac{42}{13}\\0&0&0&\vdots&0\\0&0&0&\vdots&0\end{array} \right)$

The equivalent linear system associated to this matrix is

$\begin{cases}15x+13y+4z=8\\\frac{52}{15}y+\frac{91}{15}z=-\frac{73}{15}\\2z=-\frac{42}{13}\end{cases}$

To Solve this last system is very simple by substitution. The solutions are:

$x=-\frac{7}{26}, y=\frac{37}{26}, z=-\frac{21}{13}$

7 0

2 years ago

Fill in the blank to make the two fractions equivalent.

vovikov84 [41]

The answer is 20. Its there but erased

6 0

2 years ago

Read 2 more answers