How can I solve this system of equation with substitution u+6y=32 and u+3y=17

Mathematics

2 answers:

Ne4ueva [31]3 years ago

6 0

U+6y=32 U+3y=17

Solve U+6y=32 for u

Add -6y to both sides

U+6y-6y=32-6y U= -6y+32

Substitute -6y+32 for U in U+3y=17 U+3y=17 -6y+32+3y=17

Simplify both sides of the equation -3y+32=17

Add -32 to both sides -3y+32-32=17-32 -3y=-15

Divide both sides by -3 -3y/-3 = -15/-3 Y=5

Substitute 5 for y in U=-6y+32 U= -6y+32

U= (-6)(5)+32 u= -30 +32 U= 2

Answer : U= 2 and Y=5

I hope that's help and have a great night !

gayaneshka [121]3 years ago

5 0

U= 32 - 6y
u = 17 - 3y

Step 1
32 - 6y = 17 - 3y

Step 2
-6y + 3y = 17 - 32

Step 3
-3y = - 15

Step 4
3y = 15

Step 5
y = 15/3

Step 6
y = 5

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