05.12.2021

Solve the following systems of equations by any method.
1. 2y - 4 = 0
x + 2y = 5

. 0

Step-by-step answer

24.06.2023, solved by verified expert

Faq

Mathematics
Step-by-step answer
P Answered by PhD
(1,2)
good luck and pls mark brainliest
Mathematics
Step-by-step answer
P Answered by PhD
Well, 50 points total given, 25 points per user  and 13 bonus for brainliest

anyway

1.
add them equations, the y's will cancel
x+2y=4
3x-2y=4 +
4x+0y=8

4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution

2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
-4x+14y=-6 +
0x+0y=0
0=0
infinite solutions

that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions

3.
multiply first equation by 2 and add to first
4x+2y=-6
1x-2y=-4 +
5x+0y=-10

5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1

x=-2
y=1
(x,y)
(-2,1)

4.
coincident means they are the same line
so

we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16

5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a

40 tickets, c+a
40=c+a

the equations are
150=3c+5a and
40=c+a

eliminate
multiply 2nd equaton by -3 and ad to first one

150=3c+5a
-120=-3c-3a +
30=0c+2a

30=2a
divide by 2
15=a

sub back
40=c+a
40=c+15
minus 15
25=c

25 children tickets and 15 adult tickets were sold

ANSWERS:

1.
(2,1) is solution

2.
infinite solutions

3.
(-2,1)

4.
B=10 and C=16

5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
the equations are
150=3c+5a and
40=c+a
25 children tickets and 15 adult tickets were sold
Mathematics
Step-by-step answer
P Answered by PhD
Well, 50 points total given, 25 points per user  and 13 bonus for brainliest

anyway

1.
add them equations, the y's will cancel
x+2y=4
3x-2y=4 +
4x+0y=8

4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution

2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
-4x+14y=-6 +
0x+0y=0
0=0
infinite solutions

that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions

3.
multiply first equation by 2 and add to first
4x+2y=-6
1x-2y=-4 +
5x+0y=-10

5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1

x=-2
y=1
(x,y)
(-2,1)

4.
coincident means they are the same line
so

we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16

5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a

40 tickets, c+a
40=c+a

the equations are
150=3c+5a and
40=c+a

eliminate
multiply 2nd equaton by -3 and ad to first one

150=3c+5a
-120=-3c-3a +
30=0c+2a

30=2a
divide by 2
15=a

sub back
40=c+a
40=c+15
minus 15
25=c

25 children tickets and 15 adult tickets were sold

ANSWERS:

1.
(2,1) is solution

2.
infinite solutions

3.
(-2,1)

4.
B=10 and C=16

5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
the equations are
150=3c+5a and
40=c+a
25 children tickets and 15 adult tickets were sold
Mathematics
Step-by-step answer
P Answered by PhD

We would use the first equation to solve the second.

First, rewrite x+y=300 as x=300-y

Now we can substitute that for x in the second equation, allowing us to only have to solve for y.

300-y+3y=18

300-4y=18

-4y=-282

4y=282

Now, divide both sides by 4 to isolate x.

Y= 70.5

Now that we know the value of y, insert that in our first equation.

X=300-70.5

X=229.5

Mathematics
Step-by-step answer
P Answered by PhD

We would use the first equation to solve the second.

First, rewrite x+y=300 as x=300-y

Now we can substitute that for x in the second equation, allowing us to only have to solve for y.

300-y+3y=18

300-4y=18

-4y=-282

4y=282

Now, divide both sides by 4 to isolate x.

Y= 70.5

Now that we know the value of y, insert that in our first equation.

X=300-70.5

X=229.5

Mathematics
Step-by-step answer
P Answered by PhD

x = 5 , y = 2

Step-by-step explanation:

Solve the following system:

{y = (7 x)/5 - 5 | (equation 1)

{y = (3 x)/5 - 1 | (equation 2)

Express the system in standard form:

{-(7 x)/5 + y = -5 | (equation 1)

{-(3 x)/5 + y = -1 | (equation 2)

Subtract 3/7 × (equation 1) from equation 2:

{-(7 x)/5 + y = -5 | (equation 1)

{0 x+(4 y)/7 = 8/7 | (equation 2)

Multiply equation 1 by 5:

{-(7 x) + 5 y = -25 | (equation 1)

{0 x+(4 y)/7 = 8/7 | (equation 2)

Multiply equation 2 by 7/4:

{-(7 x) + 5 y = -25 | (equation 1)

{0 x+y = 2 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(7 x)+0 y = -35 | (equation 1)

{0 x+y = 2 | (equation 2)

Divide equation 1 by -7:

{x+0 y = 5 | (equation 1)

{0 x+y = 2 | (equation 2)

Collect results:

 {x = 5 , y = 2

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