All categories

3 years ago

Solve the system of equations y=x^2+3x-4 and y=2x-4

Mathematics

1 answer:

laila [671]3 years ago

4 0

Answer:

Step-by-step explanation:

Given the 2 equations

y = x² + 3x - 4 → (1)

y = 2x - 4 → (2)

Substitute y = x² + 3x - 4 into (2)

x² + 3x - 4 = 2x - 4 ( subtract 2x - 4 from both sides )

x² + x = 0 ← in standard form

x(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x = 0

x + 1 = 0 ⇒ x = - 1

Substitute these values into (2) for corresponding values of y

x = 0 → y = 2(0) - 4 = 0 - 4 = - 4 ⇒ (0, - 4 )

x = - 1 → y = 2(- 1) - 4 = - 2 - 4 = - 6 ⇒ (- 1, - 6 )

You might be interested in

Rachel's cell phone company charges $0.10 per minute. She used 350 minutes last month and paid $84. writ and solve a linear equa

dimaraw [331]

Answer:

Step-by-step explanation:

cuz,
x equals the number of minutes so use the y slope formula

0.10(x)+the intercept = y
0.10(350)+the intercept = 84
35+the intercept = 84
the intercept = 49
so the equation is
0.10(x)+49 = y
or
1/10(x)+49 = y
1/10(300)+49=y
30+49=y
79=y
so the answer is 79 dollars

4 0

3 years ago

PLEASEEEE HELPPPPPPPPPP

natita [175]

5 the answer I took the test

8 0

3 years ago

Read 2 more answers

What is 150% of 200. Show work. ASAP

zalisa [80]

150 % of 200

150 / 100 = 1.5

1.5 * 200 = 300

hope this helps!.

8 0

3 years ago

Read 2 more answers

Help fast please....

IrinaVladis [17]

The second one is the answer

7 0

3 years ago

A certain forest covers an area of

iVinArrow [24]

Answer:

2598 square kilometers

Step-by-step explanation:

Hello

Step 1

year one

using a rule of three is possible to find how much is 8.75 od 4500 km2

Let

4500 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$4500:100\\x:8.75\\\frac{4500}{100}=\frac{x}{8.75}\\x=\frac{4500*8.75}{100} \\x=393.75\\$

at the end of the year one, the area will be

4500-393.75=4106.25

this will be the initial area for the year 2.

Step 2

repite the step 1 with area initial =4106.25 km2

4106.25 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$4106.25:100\\x:8.75\\\frac{4106.25}{100}=\frac{x}{8.75}\\x=\frac{4106.25*8.75}{100} \\x=359.29\\$

at the end of the year 2, the area will be

4106-359.29=3746.70

this will be the initial area for the year 3.

Step 3

repite the step 1 with area initial =4106.25 km2

3746.70 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3746.70:100\\x:8.75\\\frac{3746.70}{100}=\frac{x}{8.75}\\x=\frac{3746.7*8.75}{100} \\x=327.83\\$

at the end of the year 3, the area will be

3746.70-327.83=3419.09

this will be the initial area for the year 4.

Step 4

year four

repite the step 1 with area initial =3419.09 km2

3419.09 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3419.09:100\\x:8.75\\\frac{3419.09}{100}=\frac{x}{8.75}\\x=\frac{3419.09*8.75}{100} \\x=299.17\\$

at the end of the year 4, the area will be

3419.09-299.173=3119.82

this will be the initial area for the year 5.

Step 5

year five

repite the step 1 with area initial =3119.82 km2

3119.82 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3119.82:100\\x:8.75\\\frac{3119.82}{100}=\frac{x}{8.75}\\x=\frac{3119.82*8.75}{100} \\x=272.99\\$

at the end of the year 5, the area will be

3119.82-272.99=2846.92

this will be the initial area for the year 6.

Step 6

year six

repite the step 1 with area initial =2846.92km2

2846.92 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$2846.92:100\\x:8.75\\\frac{2846.92}{100}=\frac{x}{8.75}\\x=\frac{2846.92*8.75}{100} \\x=249.10\\$

at the end of the year six, the area will be

2846.92-249.10=2597.82 square kilometers

Have a great day.

8 0

3 years ago