Question

equations using linear combination
equations?\tmultiply each equation by a number that produces opposite coefficients for x or y.
\t\t4x + 5y = 7\t x
\t\t3x - 2y = -12\t x
\t\t\t-5
\t\t\t-4
\t\t\t-3
\t\t\t-2
\t\t\t-1
\t\t\t1
\t\t\t2
\t\t\t3
\t\t\t4
\t\t\t5

Explanation:

Step1: Target opposite x-coefficients

To get opposite coefficients for $x$, we multiply the first equation by $3$ and the second by $-4$ (or first by $-3$ and second by $4$).
First equation: $3\times(4x + 5y = 7) \implies 12x + 15y = 21$
Second equation: $-4\times(3x - 2y = -12) \implies -12x + 8y = 48$

Step2: Target opposite y-coefficients

To get opposite coefficients for $y$, we multiply the first equation by $2$ and the second by $5$.
First equation: $2\times(4x + 5y = 7) \implies 8x + 10y = 14$
Second equation: $5\times(3x - 2y = -12) \implies 15x - 10y = -60$

Answer:

Option 1 (for eliminating $x$): Multiply $4x + 5y = 7$ by $3$ and $3x - 2y = -12$ by $-4$ (or vice versa: multiply first by $-3$ and second by $4$)
Option 2 (for eliminating $y$): Multiply $4x + 5y = 7$ by $2$ and $3x - 2y = -12$ by $5$