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For what values of a can you solve the linear system az + 3y = 2 and 4x + 5y = 6 by elimination without
multiplying first?
The values
and

Sagot :

Mathstudent55idn

Answer:

4, -4

Step-by-step explanation:

I think the first equation is ax + 3y = 2, not az + 3y = 2.

ax + 3y = 2

4x + 5y = 6

To use elimination, you need the coefficients of a to either add to zero, or to have a difference of zero if you subtract them.

a + 4 = 0

a = -4

a - 4 = 0

a = 4

Answer: a must be 4 or -4.

Semsee45idn

Answer:

-4 and 4

Step-by-step explanation:

Given linear system:

[tex]ax+3y=2[/tex]

[tex]4x+5y=6[/tex]

In order to solve the given linear system by elimination without multiplying first, [tex]a[/tex] must be either -4 or 4.  That way, when you add or subtract the equations, the variable [tex]x[/tex] will be eliminated.

when [tex]a = -4[/tex]:

[tex]\large\begin{array}{ l r c c l r}& -4x & + & 3y & = & 2\\+ & 4x & + & 5y & = & 6\\\cline{1-6}& & & 8y & = & 8\\\cline{1-6}\end{array}[/tex]

when [tex]a = 4[/tex]:

[tex]\large\begin{array}{ l r c c l r}& 4x & + & 3y & = & 2\\- & 4x & + & 5y & = & 6\\\cline{1-6}& & & -2y & = & -4\\\cline{1-6}\end{array}[/tex]

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