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3 years ago

The x coordinate of the solution to the system of equations y=x+4 and 3y=-2x+2

Mathematics

2 answers:

djyliett [7]3 years ago

6 0

Answer:

The x coordinate of the solution is -2

Step-by-step explanation:

* To find the x-coordinate of the solution to the system of the

equations y = x + 4 and 3y = -2x + 2 solve the equations by the

substitution method

- In the substitution method we substitute one of the two variables

by the other to make an equation of one variable

∵ y = x + 4 ⇒ (1)

∵ 3y = -2x + 2 ⇒ (2)

- Substitute the value of y in equation (2) by the value of y in equation (1)

∵ The value y = 4 + x in the equation (1)

- Put this value of y in the equation (2)

∴ 3(x + 4) = -2x + 2 ⇒ open the bracket

∴ 3x + 12 = -2x + 2 ⇒ subtract 12 from both sides

∴ 3x = -2x - 10 ⇒ add 2x to both sides

∴ 5x = -10 ⇒ divide both sides by 5

∴ x = -2

* The x coordinate of the solution is -2

Vera_Pavlovna [14]3 years ago

3 0

<h2>Hello!</h2>

The answer is:

The x-coordinate of the solution to the system of equations is:

$x=-2$

We can solve the problem writing both equations as a system of equations.

So, we are given the equations:

$\left \{ {{y=x+4} \atop {3y=-2x+2}} \right.$

Then, solving by reduction we have:

Multiplying the first equation by 2 in order to reduce the variable "x", we have:

$\left \{ {{2y=2x+2*4} \atop {3y=-2x+2}} \right.$

$5y=2x-2x+8+2\\\\5y=8+2\\\\y=\frac{10}{5}=2$

Now, substituting "y" into the first equation, to isolate "x" we have:

$y=x+4\\\\2=x+4\\\\x=2-4=-2$

Hence we have that the x-coordinate of the solution to the system of equations is

$x=-2$

Have a nice day!

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Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

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Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

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step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

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$-2(x-2)^{2}=-5$

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$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

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$x=2(-)1.58=0.42$

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we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

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$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

$-(x^{2}-50x+625)=-361$

$(x^{2}-50x+625)=361$

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

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This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

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Rewrite as perfect square

$-2(x-7)^{2}+118$

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Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

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Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

The vertex is the point (4,6)

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Part 8) we have

$x^{2}+6x+5$

Convert to vertex form

Group terms

$(x^{2}+6x)+5$

Complete the square

$(x^{2}+6x+9)+5-9$

$(x^{2}+6x+9)-4$

Rewrite as perfect squares

$(x+3)^{2}-4$

Part 9) we have

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This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

$2(x^{2}-2x)-2=0$

Complete the square

$2(x^{2}-2x+1)-2-2=0$

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

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Part 10) we have

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This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

$8(x^{2}-8x)+720$

Complete the square

$8(x^{2}-8x+16)+720-128$

$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

$8(x-4)^{2}+592$

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6 0

3 years ago