Consider the first equation. Combine −3y and 5y to get 2y.
x=2y+5−3x
Subtract 2y from both sides.
x−2y=5−3x
Add 3x to both sides.
x−2y+3x=5
Combine x and 3x to get 4x.
4x−2y=5
Consider the second equation. Combine −3y and 5y to get 2y.
2y+5−3x=−3
Subtract 5 from both sides.
2y−3x=−3−5
Subtract 5 from −3 to get −8.
2y−3x=−8
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
4x−2y=5,−3x+2y=−8
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
4x−2y=5
Add 2y to both sides of the equation.
4x=2y+5
Divide both sides by 4.
x=
4
1
(2y+5)
Multiply
4
1
times 2y+5.
x=
2
1
y+
4
5
Substitute
2
y
+
4
5
for x in the other equation, −3x+2y=−8.
−3(
2
1
y+
4
5
)+2y=−8
Multiply −3 times
2
y
+
4
5
.
−
2
3
y−
4
15
+2y=−8
Add −
2
3y
to 2y.
2
1
y−
4
15
=−8
Add
4
15
to both sides of the equation.
2
1
y=−
4
17
Multiply both sides by 2.
y=−
2
17
Substitute −
2
17
for y in x=
2
1
y+
4
5
. Because the resulting equation contains only one variable, you can solve for x directly.
x=
2
1
(−
2
17
)+
4
5
Multiply
2
1
times −
2
17
by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=
4
−17+5
Add
4
5
to −
4
17
by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
[tex]\\[/tex]
[tex] \LARGE\sf\underline{\underline{\fbox\purple{\tt{SoLuTiOn}}}}:-[/tex]
⠀
⠀
[tex]\bf\implies{x +3y=5}[/tex]
⠀
[tex]\sf\implies{x=5-3y}[/tex](we subtract 3y on each side)
⠀
⠀
Now that we have [tex]{x}[/tex] we can substitute in the second equation
⠀
⠀
[tex]\sf\implies{4x+5y=13}[/tex]
⠀
[tex]\sf\implies{4.(5-3y)+5y=13}[/tex]
⠀
[tex]\sf\implies{20-12y+5y=13}[/tex]
⠀
[tex]\sf\implies{20-7y=13}[/tex]
⠀
[tex]\sf\implies{20=13+7y}[/tex] (we add 7y on each side)
⠀
[tex]\sf\implies{20-13=7y}[/tex]
⠀⠀
[tex]\implies{\tt{7y=7}}[/tex]
⠀
[tex]\implies{\tt{y=1}}[/tex] (we divide by 7 on each side)
⠀
⠀
Now that we have [tex]{y}[/tex] we can find [tex]{x}[/tex] :
⠀
⠀
[tex]\rm\implies{x=5-3y}[/tex]
⠀
[tex]\sf\implies{x=5-3.1}[/tex]
⠀
[tex]\implies{\tt{x=5-3}}[/tex]
⠀
[tex]\bf\implies{x=2}[/tex]
[tex]\\[/tex]
[tex]\\[/tex]
[tex]\LARGE\underline{\underline{\mathtt{\purple{Hope\:}\orange{it\:}\pink{Helps\:}\blue{uh\:}\red{Dear\:}\green{!!}}}}[/tex]
[tex]\\[/tex]
Answer:
x=−3y+5−3x+5y=−3
Consider the first equation. Combine −3y and 5y to get 2y.
x=2y+5−3x
Subtract 2y from both sides.
x−2y=5−3x
Add 3x to both sides.
x−2y+3x=5
Combine x and 3x to get 4x.
4x−2y=5
Consider the second equation. Combine −3y and 5y to get 2y.
2y+5−3x=−3
Subtract 5 from both sides.
2y−3x=−3−5
Subtract 5 from −3 to get −8.
2y−3x=−8
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
4x−2y=5,−3x+2y=−8
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
4x−2y=5
Add 2y to both sides of the equation.
4x=2y+5
Divide both sides by 4.
x=
4
1
(2y+5)
Multiply
4
1
times 2y+5.
x=
2
1
y+
4
5
Substitute
2
y
+
4
5
for x in the other equation, −3x+2y=−8.
−3(
2
1
y+
4
5
)+2y=−8
Multiply −3 times
2
y
+
4
5
.
−
2
3
y−
4
15
+2y=−8
Add −
2
3y
to 2y.
2
1
y−
4
15
=−8
Add
4
15
to both sides of the equation.
2
1
y=−
4
17
Multiply both sides by 2.
y=−
2
17
Substitute −
2
17
for y in x=
2
1
y+
4
5
. Because the resulting equation contains only one variable, you can solve for x directly.
x=
2
1
(−
2
17
)+
4
5
Multiply
2
1
times −
2
17
by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=
4
−17+5
Add
4
5
to −
4
17
by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=−3
The system is now solved.
x=−3,y=− 17/2
Step-by-step explanation: