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3 years ago

3. Solve using substitution. x - y = 4 3x - 2y = -5

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

6 0

<u>Given</u>:

The system of linear equations are $x-y=4$ and $3x-2y=-5$

We need to determine the solution to the system of equations using substitution method.

<u>Solution</u>:

The solution can be determined using the substitution method.

Let us substitute $x=4+y$ in the equation $3x-2y=-5$

Thus, we have;

$3(4+y)-2y=-5$

$12+3y-2y=-5$

$12+y=-5$

$y=-17$

Thus, the value of y is -17.

Substituting $y=-17$ in the equation $x-y=4$ , we get;

$x+17=4$

$x=-13$

Thus, the value of x is -13.

Hence, the solution to the system of equations is (-13,-17)

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PLEASE HELP A flight across the US takes longer east to west then it does west to east. This is due to the plane having a headwi

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To solve our questions, we are going to use the kinematic equation for distance: $d=vt$
where
$d$ is distance
$v$ is speed
$t$ is time

1. Let $v_{w}$ be the speed of the wind, $t_{w}$ be time of the westward trip, and $t_{e}$ the time of the eastward trip. We know from our problem that the distance between the cities is 2,400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/hr, so $v=450$ . Now we can use our equation the relate the unknown quantities with the quantities that we know:

<span>Going westward:
The plane is flying against the wind, so we need to subtract the speed of the wind form the speed of the plane:
</span> $d=vt$
$2400=(450-v_{w})t_{w}$

Going eastward:
The plane is flying with the wind, so we need to add the speed of the wind to the speed of the plane:
$d=vt$
$2400=(450+v_{w})t_{e}$

We can conclude that you should complete the chart as follows:
Going westward -Distance: 2400 Rate: $450-v_w$ Time: $t_w$
Going eastward -Distance: 2400 Rate: $450+v_w$ Time: $t_e$

2. Notice that we already have to equations:
Going westward: $2400=(450-v_{w})t_{w}$ equation(1)
Going eastward: $2400=(450+v_{w})t_{e}$ equation (2)

Let $t_{t}$ be the time of the round trip. We know from our problem that the round trip takes 11 hours, so $t_{t}=11$ , but we also know that the time round trip is the time of the westward trip plus the time of the eastward trip, so $t_{t}=t_w+t_e$ . Using this equation we can express $t_w$ in terms of $t_e$ :
$t_{t}=t_w+t_e$
$11=t_w+t_e$ equation
$t_w=11-t_e$ equation (3)
Now, we can replace equation (3) in equation (1) to create a system of equations with two unknowns:
$2400=(450-v_{w})t_{w}$
$2400=(450-v_{w})(11-t_e)$

We can conclude that the system of equations that represent the situation if the round trip takes 11 hours is:
$2400=(450-v_{w})(11-t_e)$ equation (1)
$2400=(450+v_{w})t_{e}$ equation (2)

3. Lets solve our system of equations to find the speed of the wind:
$2400=(450-v_{w})(11-t_e)$ equation (1)
$2400=(450+v_{w})t_{e}$ equation (2)

Step 1. Solve for $t_{e}$ in equation (2)
$2400=(450+v_{w})t_{e}$
$t_{e}= \frac{2400}{450+v_{w}}$ equation (3)

Step 2. Replace equation (3) in equation (1) and solve for $v_w$ :
$2400=(450-v_{w})(11-t_e)$
$2400=(450-v_{w})(11-\frac{2400}{450+v_{w}} )$
$2400=(450-v_{w})( \frac{4950+11v_w-2400}{450+v_{w}} )$
$2400=(450-v_{w})( \frac{255011v_w}{450+v_{w}} )$
$2400= \frac{1147500+4950v_w-2550v_w-11(v_w)^2}{450+v_{w}}$
$2400(450+v_{w})=1147500+2400v_w-11(v_w)^2$
$1080000+2400v_w=1147500+2400v_w-11(v_w)^2$
$(11v_w)^2-67500=0$
$11(v_w)^2=67500$
$(v_w)^2= \frac{67500}{11}$
$v_w= \sqrt{\frac{67500}{11}}$
$v_w=78$

We can conclude that the speed of the wind is 78 mi/hr.

6 0

3 years ago

6+8d=7d solve for d.

Margarita [4]

Question -

6 + 8d = 7d

Solution-

6. = 7d - 8d

6. = -d

d. = -6

<h3>Verification </h3>

6 + 8(-6) = 7(-6)