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hjlf
3 years ago
10

Solve the system of equations by graphing. Check the solution. y=-2x-3. y=2x+5. Graph each equation on the same coordinate plane

. At what point do the graphs of the lines appear to intersect?​
Mathematics
2 answers:
Semenov [28]3 years ago
8 0

Answer:

\displaystyle [-2, 1]

Step-by-step explanation:

Use the Elimination method sinse both coefficients in the system are OPPOCITES:

\displaystyle \left \{{{y = -2x - 3} \atop {y = 2x + 5}} \right.

2<em>y</em> = 2; 1 = <em>y</em> [Plug this y-coordinate back into the system to get the x-coordinate of −2 (−2 = <em>x</em>).]

I am joyous to assist you at any time.

sammy [17]3 years ago
7 0

Answer:

\left \{ {{x=-2} \atop {y=1}} \right.

Step-by-step explanation:

\left \{ {{y=-2x-3} \atop {y=2x+5}} \right.

<u><em>Substitute into one of the equations
</em></u>-2x-3=2x+5

<u><em>Rearrange variables to the left side of the equation
</em></u>-2x-2x=5+3

<u><em>Combine like terms
</em></u>-4x=5+3

<u><em>Calculate the sum or difference
</em></u>-4x=8

<u><em>Divide both sides of the equation by the coefficient of variable
</em></u>x=-\frac{8}{4}

<u><em>Cross out the common factor
</em></u>x=-2

<u><em>Substitute into one of the equations
</em></u>y=-2\times\left(-2\right)-3

<u><em>Calculate
</em></u>y=1

<u><em>The solution of the system is
</em></u>\left \{ {{x=-2} \atop {y=1}} \right.

<em>I hope this helps you! Don't forget to check the picture I uploaded as well</em>

<em>:)</em>

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Step-by-step explanation:

=−1−(−2)

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3 years ago
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mylen [45]

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1+2/n=3n

Step-by-step explanation:

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2 years ago
Roshonda begins riding her bike home from school at 3:00p.M., traveling 12 miles per hour. James leaves school in a bus a quarte
weeeeeb [17]

Answer:

Step-by-step explanation:

At the point where Roshonda and James meet each other, they must have covered equal distances.

Let x represent the distance that Roshonda covered while riding from school to home and also the distance covered by James while travelling in a bus from school.

Let t represent the time it took Roshonda to cover x miles.

Distance = speed × time

Therefore

x = 12 × t = 12t

James left school 0.25 hours later at a speed of 35 miles per hour. Time spent by James in covering x miles would be t - 0.25.

Distance covered would be

x = 35(t - 0.25) = 35t - 8.75

Since the distance covered is the same, then,

12t = 35t - 8.75

35t - 12t = 8.75

23t = 8.75

t = 8.75/23 = 0.38 hours

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Read 2 more answers
For what values of the variable, do the following fractions exist: y^2-1/y+y/y-3
ale4655 [162]

Answer:

Remember that the division by zero is not defined, this is the criteria that we will use in this case.

1) \frac{y^2 - 1}{y}  + \frac{y}{y - 3}

So the fractions are defined such that the denominator is never zero.

For the first fraction, the denominator is zero when y = 0

and for the second fraction, the denominator is zero when y = 3

Then the fractions exist for all real values except for y = 0 or y = 3

we can write this as:

R / {0} U { 3}

(the set of all real numbers except the elements 0 and 3)

2) \frac{b + 4}{b^2 + 7}

Let's see the values of b such that the denominator is zero:

b^2 + 7 = 0

b^2 = -7

b = √-7

This is a complex value, assuming that b can only be a real number, there is no value of b such that the denominator is zero, then the fraction is defined for every real number.

The allowed values are R, the set of all real numbers.

3) \frac{a}{a*(a - 1) - 1}

Again, we need to find the value of a such that the denominator is zero.

a*(a - 1) - 1 = a^2 - a - 1

So we need to solve:

a^2 - a - 1 = 0

We can use the Bhaskara's formula, the two values of a are given by:

a = \frac{-(-1) \pm \sqrt{(-1)^2 + 4*1*(-1)}  }{2*1}  = \frac{1 \pm \sqrt{5} }{2}

Then the two values of a that are not allowed are:

a = (1 + √5)/2

a = (1 - √5)/2

Then the allowed values of a are:

R / {(1 + √5)/2} U {(1 - √5)/2}

7 0
3 years ago
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