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Can you solve for a,b and c

2a - 3b = -5

4a + 3b = 17

50 - 2a = 1


Sagot :

Answer:

[tex]a = \frac{49}{2}[/tex]

[tex]b = 18[/tex]

Step-by-step explanation:

Lets focus on the third equation to find a:

50 - 2a = 1

Add 2a to both sides of the equation:

50 - 2a + 2a = 1 + 2a

50 = 1 + 2a

Subtract 1 from both sides of the equation:

50 - 1 = 1 + 2a - 1

49 = 2a

Divide both sides by 2:

[tex]\frac{2a}{2} = \frac{49}{2}[/tex]

[tex]a = \frac{49}{2}[/tex]

Now substitute in the know a variable into the first equation:

[tex]2a - 3b = -5[/tex]

[tex]49 - 3b = -5[/tex]

Add 3b to both sides:

[tex]49 - 3b + 3b = -5 + 3b\\49 = -5 + 3b[/tex]

Add 5 to both sides:

[tex]49 + 5 = -5 + 3b + 5[/tex]

54 = 3b

Divide both sides by 3:

[tex]\frac{53}{3} = \frac{3b}{3}[/tex]

[tex]b = 18[/tex]

c cannot be solved as there is no c variable.

Hope this helps!