Resources
Testimonials
Plans
Sign in
Sign up
Resources
Testimonials
Plans
AI tutor
Welcome to Bytelearn!
Let’s check out your problem:
Solve the system of equations.
\newline
y
=
−
14
x
−
33
y = -14x - 33
y
=
−
14
x
−
33
\newline
y
=
x
2
−
26
x
+
3
y = x^2 - 26x + 3
y
=
x
2
−
26
x
+
3
\newline
Write the coordinates in exact form. Simplify all fractions and radicals.
\newline
(______,______)
View step-by-step help
Home
Math Problems
Algebra 1
Solve a system of linear and quadratic equations
Full solution
Q.
Solve the system of equations.
\newline
y
=
−
14
x
−
33
y = -14x - 33
y
=
−
14
x
−
33
\newline
y
=
x
2
−
26
x
+
3
y = x^2 - 26x + 3
y
=
x
2
−
26
x
+
3
\newline
Write the coordinates in exact form. Simplify all fractions and radicals.
\newline
(______,______)
Set Equations Equal:
Set the two equations equal to each other since they both equal
y
y
y
.
−
14
x
−
33
=
x
2
−
26
x
+
3
-14x - 33 = x^2 - 26x + 3
−
14
x
−
33
=
x
2
−
26
x
+
3
Move Terms, Set to Zero:
Move all terms to one side to set the equation to zero.
\newline
x
2
−
26
x
+
14
x
+
3
+
33
=
0
x^2 - 26x + 14x + 3 + 33 = 0
x
2
−
26
x
+
14
x
+
3
+
33
=
0
\newline
x
2
−
12
x
+
36
=
0
x^2 - 12x + 36 = 0
x
2
−
12
x
+
36
=
0
Factor Quadratic Equation:
Factor the quadratic equation.
(
x
−
6
)
(
x
−
6
)
=
0
(x - 6)(x - 6) = 0
(
x
−
6
)
(
x
−
6
)
=
0
Solve for x:
Solve for x by setting each factor equal to zero.
\newline
x
−
6
=
0
x - 6 = 0
x
−
6
=
0
\newline
x
=
6
x = 6
x
=
6
Substitute
x
x
x
for
y
y
y
:
Substitute
x
x
x
back into one of the original equations to find
y
y
y
. Using
y
=
−
14
x
−
33
y = -14x - 33
y
=
−
14
x
−
33
, substitute
x
=
6
x = 6
x
=
6
.
y
=
−
14
(
6
)
−
33
y = -14(6) - 33
y
=
−
14
(
6
)
−
33
y
=
−
84
−
33
y = -84 - 33
y
=
−
84
−
33
y
=
−
117
y = -117
y
=
−
117
Write Coordinates:
Write the coordinates in exact form.
\newline
The solution to the system is
(
6
,
−
117
)
(6, -117)
(
6
,
−
117
)
.
More problems from Solve a system of linear and quadratic equations
Question
Solve for
v
v
v
.
\newline
v
2
+
8
v
+
12
=
0
v^2+8v+12=0
v
2
+
8
v
+
12
=
0
\newline
Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.
\newline
v
=
v=
v
=
_____
Get tutor help
Posted 1 month ago
Question
Solve for
u
u
u
.
\newline
u
2
=
9
u^2 = 9
u
2
=
9
\newline
Write your answers as integers or as proper or improper fractions in simplest form.
\newline
u
=
u =
u
=
_____ or
u
=
u =
u
=
_____
Get tutor help
Posted 2 months ago
Question
Complete the square. Fill in the number that makes the polynomial a perfect-square quadratic.
\newline
k
2
−
20
k
+
_
_
_
_
_
k^2 - 20k + \_\_\_\_\_
k
2
−
20
k
+
_____
Get tutor help
Posted 2 months ago
Question
Solve by completing the square.
\newline
u
2
−
24
u
−
23
=
0
u^2 - 24u - 23 = 0
u
2
−
24
u
−
23
=
0
\newline
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.
\newline
`u` = ________ or `u` =______
Get tutor help
Posted 2 months ago
Question
Find the discriminant.
\newline
2
q
2
−
9
q
+
8
=
0
2q^2 - 9q + 8 = 0
2
q
2
−
9
q
+
8
=
0
\newline
______
Get tutor help
Posted 2 months ago
Question
Solve the system of equations.
\newline
y
=
22
x
−
38
y = 22x - 38
y
=
22
x
−
38
\newline
y
=
x
2
+
11
x
−
14
y = x^2 + 11x - 14
y
=
x
2
+
11
x
−
14
\newline
Write the coordinates in exact form. Simplify all fractions and radicals.
\newline
(
_
,
_
)
(\_,\_)
(
_
,
_
)
\newline
(
_
,
_
)
(\_,\_)
(
_
,
_
)
Get tutor help
Posted 1 month ago
Question
Solve for
a
a
a
.
\newline
(
a
+
5
)
(
a
+
1
)
=
0
(a + 5)(a + 1) = 0
(
a
+
5
)
(
a
+
1
)
=
0
\newline
Write your answers as integers or as proper or improper fractions in simplest form.
\newline
a
=
a =
a
=
_____ or
a
=
a =
a
=
_____
Get tutor help
Posted 2 months ago
Question
Let
p
=
x
2
−
7
p=x^{2}-7
p
=
x
2
−
7
. Which equation is equivalent to
(
x
2
−
7
)
2
−
4
x
2
+
28
=
5
(x^{2}-7)^{2}-4x^{2}+28=5
(
x
2
−
7
)
2
−
4
x
2
+
28
=
5
in terms of
p
p
p
? Choose
1
1
1
answer:
\newline
(A)
p
2
−
4
p
+
23
=
0
p^{2}-4p+23=0
p
2
−
4
p
+
23
=
0
\newline
(B)
p
2
+
4
p
−
5
=
0
p^{2}+4p-5=0
p
2
+
4
p
−
5
=
0
\newline
(C)
p
2
−
4
p
−
5
=
0
p^{2}-4p-5=0
p
2
−
4
p
−
5
=
0
\newline
(D)
p
2
+
4
p
+
23
=
0
p^{2}+4p+23=0
p
2
+
4
p
+
23
=
0
Get tutor help
Posted 2 months ago
Question
Let
m
=
2
x
+
3
m=2x+3
m
=
2
x
+
3
. Which equation is equivalent to
(
2
x
+
3
)
2
−
14
x
−
21
=
−
6
(2x+3)^{2}-14x-21=-6
(
2
x
+
3
)
2
−
14
x
−
21
=
−
6
in terms of
m
m
m
? Choose
1
1
1
answer:
\newline
(A)
m
2
+
7
m
+
6
=
0
m^{2}+7m+6=0
m
2
+
7
m
+
6
=
0
\newline
(B)
m
2
−
7
m
−
15
=
0
m^{2}-7m-15=0
m
2
−
7
m
−
15
=
0
\newline
(C)
m
2
−
7
m
+
6
=
0
m^{2}-7m+6=0
m
2
−
7
m
+
6
=
0
\newline
(D)
m
2
+
7
m
−
15
=
0
m^{2}+7m-15=0
m
2
+
7
m
−
15
=
0
Get tutor help
Posted 2 months ago
Question
Let
m
=
x
2
+
3
m=x^{2}+3
m
=
x
2
+
3
. Which equation is equivalent to
(
x
2
+
3
)
2
+
7
x
2
+
21
=
−
10
(x^{2}+3)^{2}+7x^{2}+21=-10
(
x
2
+
3
)
2
+
7
x
2
+
21
=
−
10
in terms of
m
m
m
? Choose
1
1
1
answer:
\newline
(A)
m
2
−
7
m
+
31
=
0
m^{2}-7m+31=0
m
2
−
7
m
+
31
=
0
\newline
(B)
m
2
+
7
m
+
31
=
0
m^{2}+7m+31=0
m
2
+
7
m
+
31
=
0
\newline
(C)
m
2
−
7
m
+
10
=
0
m^{2}-7m+10=0
m
2
−
7
m
+
10
=
0
\newline
(D)
m
2
+
7
m
+
10
=
0
m^{2}+7m+10=0
m
2
+
7
m
+
10
=
0
Get tutor help
Posted 2 months ago