Explain.
Label. Step1, step 2, ect.
3(x+y)^2-13(x+y)+4
Step1
substitute x+y by 'a'
given expression
=3a^2-13a+4
Step 2 Factorise
=3a^2-12a-a+4
=3a(a-4)-1(a-4)
=(a-4)(3a-1)
Step 3
Putting back the value of a
(x+y-4)(3x+3y-1) ans
1) let u = x+y
2) equation is now 3 u^2 - 13 u + 4
3) = (3u - 1) ( u - 4)
is this equation is equal to 0
then 4) 3 u = 1 and u = 4 are the possible answers
or u = 1/3 and u = 4
4) u = x + y so
x + y = 1/3 or x + y = 4
= [3(x+y) - 1][(x+y) - 4]
(x+y-4)(3x+3y-1)
Let x+y = a
3a^2 - 13a + 4
(3a -1)(a - 4)
Subst
[3(x+y) - 1][(x+y) - 4]
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Answers & Comments
Verified answer
3(x+y)^2-13(x+y)+4
Step1
substitute x+y by 'a'
given expression
=3a^2-13a+4
Step 2 Factorise
=3a^2-12a-a+4
=3a(a-4)-1(a-4)
=(a-4)(3a-1)
Step 3
Putting back the value of a
(x+y-4)(3x+3y-1) ans
1) let u = x+y
2) equation is now 3 u^2 - 13 u + 4
3) = (3u - 1) ( u - 4)
is this equation is equal to 0
then 4) 3 u = 1 and u = 4 are the possible answers
or u = 1/3 and u = 4
4) u = x + y so
x + y = 1/3 or x + y = 4
3(x+y)^2-13(x+y)+4
= [3(x+y) - 1][(x+y) - 4]
3(x+y)^2-13(x+y)+4
substitute x+y by 'a'
given expression
=3a^2-13a+4
Step 2 Factorise
=3a^2-12a-a+4
=3a(a-4)-1(a-4)
=(a-4)(3a-1)
Step 3
Putting back the value of a
(x+y-4)(3x+3y-1)
Let x+y = a
3a^2 - 13a + 4
(3a -1)(a - 4)
Subst
[3(x+y) - 1][(x+y) - 4]